Šis straipsnis yra apie iracionaliųjų funkcijų integravimą .
∫
x
x
3
4
+
1
d
x
=
∫
u
2
⋅
4
u
3
d
u
u
3
+
1
=
4
[
∫
u
2
d
u
−
∫
u
2
d
u
u
3
+
1
]
=
4
[
u
3
3
−
1
3
∫
d
(
u
3
+
1
)
u
3
+
1
]
=
{\displaystyle \int {\frac {\sqrt {x}}{x^{\frac {3}{4}}+1}}dx=\int {\frac {u^{2}\cdot 4u^{3}du}{u^{3}+1}}=4[\int u^{2}du-\int {\frac {u^{2}du}{u^{3}+1}}]=4[{\frac {u^{3}}{3}}-{\frac {1}{3}}\int {\frac {d(u^{3}+1)}{u^{3}+1}}]=}
=
4
3
[
u
3
−
ln
(
u
3
+
1
)
]
+
C
=
4
3
[
x
3
4
−
ln
(
x
3
4
+
1
)
]
+
C
,
{\displaystyle ={\frac {4}{3}}[u^{3}-\ln(u^{3}+1)]+C={\frac {4}{3}}[x^{\frac {3}{4}}-\ln(x^{\frac {3}{4}}+1)]+C,}
x
=
u
4
;
{\displaystyle x=u^{4};}
d
x
=
4
u
3
d
u
.
{\displaystyle dx=4u^{3}du.}
∫
x
+
x
2
3
+
x
1
6
x
(
1
+
x
1
3
)
d
x
=
6
∫
(
t
6
+
t
4
+
t
)
t
5
t
6
(
1
+
t
2
)
d
t
=
6
∫
t
5
+
t
3
+
1
t
2
+
1
d
t
=
6
∫
t
3
d
t
+
6
∫
d
t
t
2
+
1
=
{\displaystyle \int {\frac {x+x^{\frac {2}{3}}+x^{\frac {1}{6}}}{x(1+x^{\frac {1}{3}})}}dx=6\int {\frac {(t^{6}+t^{4}+t)t^{5}}{t^{6}(1+t^{2})}}dt=6\int {\frac {t^{5}+t^{3}+1}{t^{2}+1}}dt=6\int t^{3}dt+6\int {\frac {dt}{t^{2}+1}}=}
=
3
2
t
4
+
6
arctan
t
+
C
=
3
2
x
2
3
+
6
arctan
x
1
6
+
C
,
{\displaystyle ={\frac {3}{2}}t^{4}+6\arctan t+C={\frac {3}{2}}x^{\frac {2}{3}}+6\arctan x^{\frac {1}{6}}+C,}
x
=
t
6
;
{\displaystyle x=t^{6};}
d
x
=
6
t
5
d
t
.
{\displaystyle dx=6t^{5}dt.}
∫
2
(
2
−
x
)
2
(
2
−
x
2
+
x
)
1
3
d
x
=
−
∫
2
(
1
+
t
3
)
2
t
⋅
12
t
2
16
t
6
(
1
+
t
3
)
2
d
t
=
−
3
2
∫
d
t
t
3
=
3
4
t
2
+
C
=
3
4
(
2
+
x
2
−
x
)
2
3
+
C
,
{\displaystyle \int {\frac {2}{(2-x)^{2}}}({\frac {2-x}{2+x}})^{\frac {1}{3}}dx=-\int {\frac {2(1+t^{3})^{2}t\cdot 12t^{2}}{16t^{6}(1+t^{3})^{2}}}dt=-{\frac {3}{2}}\int {\frac {dt}{t^{3}}}={\frac {3}{4t^{2}}}+C={\frac {3}{4}}({\frac {2+x}{2-x}})^{\frac {2}{3}}+C,}
2
−
x
2
+
x
=
t
3
;
{\displaystyle {\frac {2-x}{2+x}}=t^{3};}
x
=
2
−
t
3
1
+
t
3
;
{\displaystyle x={\frac {2-t^{3}}{1+t^{3}}};}
2
−
x
=
4
t
3
1
+
t
3
;
{\displaystyle 2-x={\frac {4t^{3}}{1+t^{3}}};}
−
d
x
=
12
t
2
(
1
+
t
3
)
−
4
t
3
⋅
3
t
2
(
1
+
t
3
)
2
d
t
=
12
t
2
+
12
t
5
−
12
t
5
(
1
+
t
3
)
2
d
t
=
12
t
2
(
1
+
t
3
)
2
d
t
;
{\displaystyle -dx={12t^{2}(1+t^{3})-4t^{3}\cdot 3t^{2} \over (1+t^{3})^{2}}dt={12t^{2}+12t^{5}-12t^{5} \over (1+t^{3})^{2}}dt={12t^{2} \over (1+t^{3})^{2}}dt;}
d
x
=
−
12
t
2
(
1
+
t
3
)
2
d
t
.
{\displaystyle dx={\frac {-12t^{2}}{(1+t^{3})^{2}}}dt.}
∫
d
x
(
(
x
−
1
)
3
(
x
+
2
)
5
)
1
4
=
−
∫
(
t
4
−
1
)
(
t
4
−
1
)
12
t
3
d
t
3
⋅
3
t
4
t
(
t
4
−
1
)
2
=
−
4
3
∫
d
t
t
2
=
4
3
t
+
C
=
4
3
(
x
−
1
x
+
2
)
1
4
+
C
,
{\displaystyle \int {\frac {dx}{((x-1)^{3}(x+2)^{5})^{\frac {1}{4}}}}=-\int {\frac {(t^{4}-1)(t^{4}-1)12t^{3}dt}{3\cdot 3t^{4}t(t^{4}-1)^{2}}}=-{\frac {4}{3}}\int {\frac {dt}{t^{2}}}={\frac {4}{3t}}+C={\frac {4}{3}}({\frac {x-1}{x+2}})^{\frac {1}{4}}+C,}
(
(
x
−
1
)
3
(
x
+
2
)
5
)
1
4
=
(
x
−
1
)
(
x
+
2
)
(
x
+
2
x
−
1
)
1
4
;
{\displaystyle ((x-1)^{3}(x+2)^{5})^{\frac {1}{4}}=(x-1)(x+2)({\frac {x+2}{x-1}})^{\frac {1}{4}};}
x
+
2
x
−
1
=
t
4
;
{\displaystyle {\frac {x+2}{x-1}}=t^{4};}
x
=
t
4
+
2
t
4
−
1
;
{\displaystyle x={\frac {t^{4}+2}{t^{4}-1}};}
x
−
1
=
3
t
4
−
1
;
{\displaystyle x-1={\frac {3}{t^{4}-1}};}
x
+
2
=
3
t
4
t
4
−
1
;
{\displaystyle x+2={\frac {3t^{4}}{t^{4}-1}};}
d
x
=
−
12
t
3
(
t
4
−
1
)
2
d
t
.
{\displaystyle dx={\frac {-12t^{3}}{(t^{4}-1)^{2}}}dt.}
∫
1
(
1
−
x
)
2
(
1
+
x
1
−
x
)
1
3
d
x
=
∫
(
1
+
u
3
)
2
u
⋅
6
u
2
4
(
1
+
u
3
)
2
d
u
=
3
2
⋅
u
4
4
+
C
=
3
8
⋅
1
+
x
1
−
x
(
1
+
x
1
−
x
)
1
3
+
C
,
{\displaystyle \int {\frac {1}{(1-x)^{2}}}({\frac {1+x}{1-x}})^{\frac {1}{3}}dx=\int {\frac {(1+u^{3})^{2}u\cdot 6u^{2}}{4(1+u^{3})^{2}}}du={\frac {3}{2}}\cdot {\frac {u^{4}}{4}}+C={\frac {3}{8}}\cdot {\frac {1+x}{1-x}}({\frac {1+x}{1-x}})^{\frac {1}{3}}+C,}
1
+
x
1
−
x
=
u
3
;
{\displaystyle {\frac {1+x}{1-x}}=u^{3};}
x
=
u
3
−
1
u
3
+
1
;
{\displaystyle x={\frac {u^{3}-1}{u^{3}+1}};}
d
x
=
6
u
2
(
1
+
u
3
)
2
d
u
;
{\displaystyle dx={\frac {6u^{2}}{(1+u^{3})^{2}}}du;}
1
−
x
=
2
1
+
u
3
.
{\displaystyle 1-x={\frac {2}{1+u^{3}}}.}
Oilerio keitiniai
keisti
Funkcijos
R
(
x
,
a
x
2
+
b
x
+
c
)
(
7.66
)
{\displaystyle R(x,\;{\sqrt {ax^{2}+bx+c}})\quad (7.66)}
(a, b ir c - realieji skačiai) integralas yra elementarioji funkcija.
a
x
2
+
b
x
+
c
=
u
±
x
a
{\displaystyle {\sqrt {ax^{2}+bx+c}}=u\pm x{\sqrt {a}}}
u
=
a
x
2
+
b
x
+
c
+
x
a
{\displaystyle u={\sqrt {ax^{2}+bx+c}}+x{\sqrt {a}}}
a
x
2
+
b
x
+
c
=
u
−
x
a
{\displaystyle {\sqrt {ax^{2}+bx+c}}=u-x{\sqrt {a}}}
b
x
+
c
=
u
2
−
2
a
u
x
,
{\displaystyle bx+c=u^{2}-2{\sqrt {a}}ux,}
x
=
u
2
−
c
2
a
u
+
b
,
{\displaystyle x={\frac {u^{2}-c}{2{\sqrt {a}}u+b}},}
a
x
2
+
b
x
+
c
=
u
−
u
2
−
c
2
a
u
+
b
⋅
a
=
u
(
2
a
u
+
b
)
−
(
a
u
2
−
a
c
)
2
a
u
+
b
=
a
u
2
+
b
u
+
c
a
2
a
u
+
b
,
{\displaystyle {\sqrt {ax^{2}+bx+c}}=u-{\frac {u^{2}-c}{2{\sqrt {a}}u+b}}\cdot {\sqrt {a}}={\frac {u(2{\sqrt {a}}u+b)-({\sqrt {a}}u^{2}-{\sqrt {a}}c)}{2{\sqrt {a}}u+b}}={\frac {{\sqrt {a}}u^{2}+bu+c{\sqrt {a}}}{2{\sqrt {a}}u+b}},}
d
x
=
(
u
2
−
c
2
a
u
+
b
)
′
=
2
u
⋅
(
2
a
u
+
b
)
−
(
u
2
−
c
)
⋅
2
a
(
2
a
u
+
b
)
2
d
u
=
2
a
u
2
+
b
u
+
c
a
(
2
a
u
+
b
)
2
d
u
.
{\displaystyle dx=({\frac {u^{2}-c}{2{\sqrt {a}}u+b}})'={\frac {2u\cdot (2{\sqrt {a}}u+b)-(u^{2}-c)\cdot 2{\sqrt {a}}}{(2{\sqrt {a}}u+b)^{2}}}du=2{\frac {{\sqrt {a}}u^{2}+bu+c{\sqrt {a}}}{(2{\sqrt {a}}u+b)^{2}}}\;du.}
Tinka, kai kvadrainis trinaris turi menamas šaknis.
a
x
2
+
b
x
+
c
=
x
t
±
c
{\displaystyle {\sqrt {ax^{2}+bx+c}}=xt\pm {\sqrt {c}}}
a
x
2
+
b
x
+
c
=
x
2
t
2
+
2
x
t
c
+
c
.
{\displaystyle ax^{2}+bx+c=x^{2}t^{2}+2xt{\sqrt {c}}+c.}
(Mes apsisprendėme prieš šaknį pasirinkti pliuso ženklą.) Iš čia x yra kaip racionali funkcija nuo t :
a
x
2
−
x
2
t
2
+
b
x
−
2
x
t
c
=
0
,
{\displaystyle ax^{2}-x^{2}t^{2}+bx-2xt{\sqrt {c}}=0,}
(
a
−
t
2
)
x
2
+
(
b
−
2
t
c
)
x
=
0
,
{\displaystyle (a-t^{2})x^{2}+(b-2t{\sqrt {c}})x=0,}
x
(
(
a
−
t
2
)
x
+
b
−
2
t
c
)
=
0
;
{\displaystyle x((a-t^{2})x+b-2t{\sqrt {c}})=0;}
x =0 arba
(
a
−
t
2
)
x
+
b
−
2
t
c
=
0
;
{\displaystyle (a-t^{2})x+b-2t{\sqrt {c}}=0;}
x per t , todėl x =0 netinka; sprendžiame toliau:
(
a
−
t
2
)
x
+
b
−
2
t
c
=
0
;
{\displaystyle (a-t^{2})x+b-2t{\sqrt {c}}=0;}
(
a
−
t
2
)
x
=
2
c
t
−
b
;
{\displaystyle (a-t^{2})x=2{\sqrt {c}}t-b;}
x
=
2
c
t
−
b
a
−
t
2
.
{\displaystyle x={\frac {2{\sqrt {c}}t-b}{a-t^{2}}}.}
Turime
a
x
2
+
b
x
+
c
=
x
t
+
c
=
2
c
t
−
b
a
−
t
2
⋅
t
+
c
=
2
c
t
2
−
b
t
a
−
t
2
+
c
;
{\displaystyle {\sqrt {ax^{2}+bx+c}}=xt+{\sqrt {c}}={\frac {2{\sqrt {c}}t-b}{a-t^{2}}}\cdot t+{\sqrt {c}}={\frac {2{\sqrt {c}}t^{2}-bt}{a-t^{2}}}+{\sqrt {c}};}
(
2
c
t
−
b
a
−
t
2
)
′
=
(
2
c
t
−
b
)
′
(
a
−
t
2
)
−
(
2
c
t
−
b
)
(
a
−
t
2
)
′
(
a
−
t
2
)
2
=
2
c
(
a
−
t
2
)
−
(
2
c
t
−
b
)
(
−
2
t
)
(
a
−
t
2
)
2
=
2
a
c
−
2
c
t
2
−
(
−
4
c
t
2
+
2
b
t
)
(
a
−
t
2
)
2
=
{\displaystyle \left({\frac {2{\sqrt {c}}t-b}{a-t^{2}}}\right)'={\frac {(2{\sqrt {c}}t-b)'(a-t^{2})-(2{\sqrt {c}}t-b)(a-t^{2})'}{(a-t^{2})^{2}}}={\frac {2{\sqrt {c}}(a-t^{2})-(2{\sqrt {c}}t-b)(-2t)}{(a-t^{2})^{2}}}={\frac {2a{\sqrt {c}}-2{\sqrt {c}}t^{2}-(-4{\sqrt {c}}t^{2}+2bt)}{(a-t^{2})^{2}}}=}
=
2
a
c
−
2
c
t
2
+
4
c
t
2
−
2
b
t
(
a
−
t
2
)
2
=
(
4
c
−
2
c
)
t
2
−
2
b
t
+
2
a
c
(
a
−
t
2
)
2
.
{\displaystyle ={\frac {2a{\sqrt {c}}-2{\sqrt {c}}t^{2}+4{\sqrt {c}}t^{2}-2bt}{(a-t^{2})^{2}}}={\frac {(4{\sqrt {c}}-2{\sqrt {c}})t^{2}-2bt+2a{\sqrt {c}}}{(a-t^{2})^{2}}}.}
d
x
=
2
c
t
2
−
2
b
t
+
2
a
c
(
a
−
t
2
)
2
d
t
.
{\displaystyle dx={\frac {2{\sqrt {c}}t^{2}-2bt+2a{\sqrt {c}}}{(a-t^{2})^{2}}}dt.}
a
(
x
−
x
1
)
(
x
−
x
2
)
=
u
(
x
−
x
1
)
,
{\displaystyle {\sqrt {a(x-x_{1})(x-x_{2})}}=u(x-x_{1}),}
x
1
{\displaystyle x_{1}}
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
a
x
2
+
b
x
+
c
=
a
(
x
−
x
1
)
(
x
−
x
2
)
{\displaystyle {\sqrt {ax^{2}+bx+c}}={\sqrt {a(x-x_{1})(x-x_{2})}}}
Tegu
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
a
x
2
+
b
x
+
c
.
{\displaystyle ax^{2}+bx+c.}
a
x
2
+
b
x
+
c
=
(
x
−
α
)
t
.
{\displaystyle {\sqrt {ax^{2}+bx+c}}=(x-\alpha )t.}
Kadangi
a
x
2
+
b
x
+
c
=
a
(
x
−
α
)
(
x
−
β
)
,
{\displaystyle ax^{2}+bx+c=a(x-\alpha )(x-\beta ),}
a
(
x
−
α
)
(
x
−
β
)
=
(
x
−
α
)
t
,
{\displaystyle {\sqrt {a(x-\alpha )(x-\beta )}}=(x-\alpha )t,}
a
(
x
−
α
)
(
x
−
β
)
=
(
x
−
α
)
2
t
2
,
{\displaystyle a(x-\alpha )(x-\beta )=(x-\alpha )^{2}t^{2},}
a
(
x
−
β
)
=
(
x
−
α
)
t
2
.
{\displaystyle a(x-\beta )=(x-\alpha )t^{2}.}
Iš čia randame x kaip racionalią funkciją nuo t :
a
x
−
a
β
=
x
t
2
−
α
t
2
,
{\displaystyle ax-a\beta =xt^{2}-\alpha t^{2},}
a
x
−
x
t
2
=
a
β
−
α
t
2
,
{\displaystyle ax-xt^{2}=a\beta -\alpha t^{2},}
(
a
−
t
2
)
x
=
a
β
−
α
t
2
,
{\displaystyle (a-t^{2})x=a\beta -\alpha t^{2},}
x
=
a
β
−
α
t
2
a
−
t
2
.
{\displaystyle x={\frac {a\beta -\alpha t^{2}}{a-t^{2}}}.}
Trečias Oilerio keitinys tinka kai a >0 ir kai a <0. Butina tik, kad turėtų daugianaris
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
(
a
β
−
α
t
2
a
−
t
2
)
′
=
−
2
α
t
(
a
−
t
2
)
−
(
a
β
−
α
t
2
)
(
−
2
t
)
(
a
−
t
2
)
2
=
−
2
α
a
t
+
2
α
t
3
−
(
−
2
a
β
t
+
2
α
t
3
)
(
a
−
t
2
)
2
=
{\displaystyle ({\frac {a\beta -\alpha t^{2}}{a-t^{2}}})'={\frac {-2\alpha t(a-t^{2})-(a\beta -\alpha t^{2})(-2t)}{(a-t^{2})^{2}}}={\frac {-2\alpha at+2\alpha t^{3}-(-2a\beta t+2\alpha t^{3})}{(a-t^{2})^{2}}}=}
=
−
2
α
a
t
+
2
α
t
3
+
2
a
β
t
−
2
α
t
3
(
a
−
t
2
)
2
=
−
2
α
a
t
+
2
a
β
t
(
a
−
t
2
)
2
=
(
2
a
β
−
2
a
α
)
t
(
a
−
t
2
)
2
.
{\displaystyle ={\frac {-2\alpha at+2\alpha t^{3}+2a\beta t-2\alpha t^{3}}{(a-t^{2})^{2}}}={\frac {-2\alpha at+2a\beta t}{(a-t^{2})^{2}}}={\frac {(2a\beta -2a\alpha )t}{(a-t^{2})^{2}}}.}
d
x
=
2
a
(
β
−
α
)
t
(
a
−
t
2
)
2
d
t
.
{\displaystyle dx={\frac {2a(\beta -\alpha )t}{(a-t^{2})^{2}}}dt.}
Pirmojo Oilerio keitinio pavyzdžiai
∫
d
x
5
x
2
+
x
+
1
;
{\displaystyle \int {\frac {dx}{\sqrt {5x^{2}+x+1}}};}
5
x
2
+
x
+
1
=
u
+
x
5
.
{\displaystyle {\sqrt {5x^{2}+x+1}}=u+x{\sqrt {5}}.}
5
x
2
+
x
+
1
=
u
2
+
2
u
x
5
+
5
x
2
;
x
+
1
=
u
2
+
2
5
u
x
;
x
(
1
−
2
u
5
)
=
u
2
−
1
;
x
=
u
2
−
1
1
−
2
5
u
;
{\displaystyle 5x^{2}+x+1=u^{2}+2ux{\sqrt {5}}+5x^{2};\;x+1=u^{2}+2{\sqrt {5}}ux;\;x(1-2u{\sqrt {5}})=u^{2}-1;\;x={\frac {u^{2}-1}{1-2{\sqrt {5}}u}};}
d
x
=
(
u
2
−
1
1
−
2
5
u
)
′
=
(
u
2
−
1
)
′
(
1
−
2
5
u
)
−
(
u
2
−
1
)
(
1
−
2
5
u
)
′
(
1
−
2
5
u
)
2
=
{\displaystyle dx=({\frac {u^{2}-1}{1-2{\sqrt {5}}u}})'={\frac {(u^{2}-1)'(1-2{\sqrt {5}}u)-(u^{2}-1)(1-2{\sqrt {5}}u)'}{(1-2{\sqrt {5}}u)^{2}}}=}
=
2
u
(
1
−
2
5
u
)
−
(
u
2
−
1
)
(
−
2
5
)
1
−
4
5
u
+
4
⋅
5
⋅
u
2
d
u
=
2
u
−
4
5
u
2
+
2
5
(
u
2
−
1
)
1
−
4
5
u
+
20
u
2
d
u
=
{\displaystyle ={\frac {2u(1-2{\sqrt {5}}u)-(u^{2}-1)(-2{\sqrt {5}})}{1-4{\sqrt {5}}u+4\cdot 5\cdot u^{2}}}du={\frac {2u-4{\sqrt {5}}u^{2}+2{\sqrt {5}}(u^{2}-1)}{1-4{\sqrt {5}}u+20u^{2}}}du=}
=
2
u
−
4
5
u
2
+
2
5
⋅
u
2
−
2
5
1
−
4
5
u
+
20
u
2
d
u
=
2
u
−
2
5
u
2
−
2
5
1
−
4
5
u
+
20
u
2
d
u
=
2
(
u
−
5
u
2
−
5
)
(
1
−
2
5
⋅
u
)
2
d
u
.
{\displaystyle ={\frac {2u-4{\sqrt {5}}u^{2}+2{\sqrt {5}}\cdot u^{2}-2{\sqrt {5}}}{1-4{\sqrt {5}}u+20u^{2}}}du={\frac {2u-2{\sqrt {5}}u^{2}-2{\sqrt {5}}}{1-4{\sqrt {5}}u+20u^{2}}}du={\frac {2(u-{\sqrt {5}}u^{2}-{\sqrt {5}})}{(1-2{\sqrt {5}}\cdot u)^{2}}}du.}
5
x
2
+
x
+
1
=
u
+
x
5
=
u
+
u
2
−
1
1
−
2
5
u
⋅
5
=
u
+
5
u
2
−
5
1
−
2
5
u
=
u
(
1
−
2
5
u
)
+
5
u
2
−
5
1
−
2
5
u
=
{\displaystyle {\sqrt {5x^{2}+x+1}}=u+x{\sqrt {5}}=u+{\frac {u^{2}-1}{1-2{\sqrt {5}}u}}\cdot {\sqrt {5}}=u+{\frac {{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}={\frac {u(1-2{\sqrt {5}}u)+{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}=}
=
u
−
2
5
u
2
+
5
u
2
−
5
1
−
2
5
u
=
u
−
5
u
2
−
5
1
−
2
5
u
.
{\displaystyle ={\frac {u-2{\sqrt {5}}u^{2}+{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}={\frac {u-{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}.}
∫
d
x
5
x
2
+
x
+
1
=
∫
2
(
u
−
5
u
2
−
5
)
(
1
−
2
5
⋅
u
)
2
d
t
u
−
5
u
2
−
5
1
−
2
5
u
=
2
∫
d
u
1
−
2
5
u
=
−
2
2
5
∫
d
(
1
−
u
2
5
)
1
−
2
5
u
=
{\displaystyle \int {\frac {dx}{\sqrt {5x^{2}+x+1}}}=\int {\frac {{\frac {2(u-{\sqrt {5}}u^{2}-{\sqrt {5}})}{(1-2{\sqrt {5}}\cdot u)^{2}}}dt}{\frac {u-{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}}=2\int {\frac {du}{1-2{\sqrt {5}}u}}=-{\frac {2}{2{\sqrt {5}}}}\int {\frac {d(1-u2{\sqrt {5}})}{1-2{\sqrt {5}}u}}=}
=
−
1
5
ln
|
1
−
2
5
u
|
+
C
=
−
1
5
ln
|
1
−
2
5
(
5
x
2
+
x
+
1
−
5
x
)
|
+
C
=
{\displaystyle =-{\frac {1}{\sqrt {5}}}\ln |1-2{\sqrt {5}}u|+C=-{\frac {1}{\sqrt {5}}}\ln |1-2{\sqrt {5}}({\sqrt {5x^{2}+x+1}}-{\sqrt {5}}x)|+C=}
=
−
1
5
ln
|
1
−
100
x
2
+
20
x
+
20
+
10
x
|
+
C
.
{\displaystyle =-{\frac {1}{\sqrt {5}}}\ln |1-{\sqrt {100x^{2}+20x+20}}+10x|+C.}
∫
d
x
x
2
+
3
.
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3}}}.}
x
2
+
3
=
−
x
+
t
=
t
−
x
.
{\displaystyle {\sqrt {x^{2}+3}}=-x+t=t-x.}
x
2
+
3
=
(
−
x
+
t
)
2
=
x
2
−
2
t
x
+
t
2
{\displaystyle x^{2}+3=(-x+t)^{2}=x^{2}-2tx+t^{2}}
3
=
t
2
−
2
t
x
{\displaystyle 3=t^{2}-2tx}
2
t
x
=
t
2
−
3
{\displaystyle 2tx=t^{2}-3}
x
=
t
2
−
3
2
t
.
{\displaystyle x={\frac {t^{2}-3}{2t}}.}
d
x
=
(
t
2
−
3
2
t
)
′
=
(
t
2
−
3
)
′
⋅
2
t
−
(
t
2
−
3
)
⋅
(
2
t
)
′
(
2
t
)
2
=
2
t
⋅
2
t
−
(
t
2
−
3
)
⋅
2
4
t
2
d
t
=
2
t
2
−
(
t
2
−
3
)
2
t
2
d
t
=
t
2
+
3
2
t
2
d
t
.
{\displaystyle dx=({\frac {t^{2}-3}{2t}})'={\frac {(t^{2}-3)'\cdot 2t-(t^{2}-3)\cdot (2t)'}{(2t)^{2}}}={\frac {2t\cdot 2t-(t^{2}-3)\cdot 2}{4t^{2}}}dt={\frac {2t^{2}-(t^{2}-3)}{2t^{2}}}dt={\frac {t^{2}+3}{2t^{2}}}dt.}
x
2
+
3
=
−
x
+
t
=
−
t
2
−
3
2
t
+
t
=
−
(
t
2
−
3
)
+
2
t
2
2
t
=
t
2
+
3
2
t
.
{\displaystyle {\sqrt {x^{2}+3}}=-x+t=-{\frac {t^{2}-3}{2t}}+t={\frac {-(t^{2}-3)+2t^{2}}{2t}}={\frac {t^{2}+3}{2t}}.}
∫
d
x
x
2
+
3
=
∫
t
2
+
3
2
t
2
d
t
t
2
+
3
2
t
=
∫
d
t
t
=
ln
|
t
|
+
C
=
ln
|
x
+
x
2
+
3
|
+
C
.
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3}}}=\int {\frac {{\frac {t^{2}+3}{2t^{2}}}dt}{\frac {t^{2}+3}{2t}}}=\int {\frac {dt}{t}}=\ln |t|+C=\ln |x+{\sqrt {x^{2}+3}}|+C.}
Apskaičiuoti
∫
d
x
x
+
x
2
+
x
+
1
.
{\displaystyle \int {\frac {dx}{x+{\sqrt {x^{2}+x+1}}}}.}
Sprendimas . Kadangi trinaris
x
2
+
x
+
1
{\displaystyle x^{2}+x+1}
x
2
+
x
+
1
=
t
−
x
.
{\displaystyle {\sqrt {x^{2}+x+1}}=t-x.}
x
2
+
x
+
1
=
t
2
−
2
t
x
+
x
2
{\displaystyle x^{2}+x+1=t^{2}-2tx+x^{2}}
x
+
1
=
t
2
−
2
t
x
{\displaystyle x+1=t^{2}-2tx}
x
=
t
2
−
1
1
+
2
t
{\displaystyle x={\frac {t^{2}-1}{1+2t}}}
d
x
=
(
t
2
−
1
1
+
2
t
)
′
=
2
t
(
1
+
2
t
)
−
(
t
2
−
1
)
⋅
2
(
1
+
2
t
)
2
=
2
t
+
4
t
2
−
2
t
2
+
2
(
1
+
2
t
)
2
=
2
t
+
2
t
2
+
2
(
1
+
2
t
)
2
.
{\displaystyle dx=({\frac {t^{2}-1}{1+2t}})'={\frac {2t(1+2t)-(t^{2}-1)\cdot 2}{(1+2t)^{2}}}={\frac {2t+4t^{2}-2t^{2}+2}{(1+2t)^{2}}}={\frac {2t+2t^{2}+2}{(1+2t)^{2}}}.}
Tada
∫
d
x
x
+
x
2
+
x
+
1
=
∫
2
t
+
2
t
2
+
2
t
(
1
+
2
t
)
2
d
t
.
{\displaystyle \int {\frac {dx}{x+{\sqrt {x^{2}+x+1}}}}=\int {\frac {2t+2t^{2}+2}{t(1+2t)^{2}}}dt.}
Toliau, turime
2
t
2
+
2
t
+
2
t
(
1
+
2
t
)
2
=
A
t
+
B
1
+
2
t
+
D
(
1
+
2
t
)
2
.
{\displaystyle {\frac {2t^{2}+2t+2}{t(1+2t)^{2}}}={\frac {A}{t}}+{\frac {B}{1+2t}}+{\frac {D}{(1+2t)^{2}}}.}
Padauginę abi dalis lygybės su
t
(
1
+
2
t
)
2
,
{\displaystyle t(1+2t)^{2},}
2
t
2
+
2
t
+
2
=
A
(
1
+
2
t
)
2
+
B
t
(
1
+
2
t
)
+
D
t
=
A
(
1
+
4
t
+
4
t
2
)
+
B
t
+
2
B
t
2
+
D
t
=
(
4
A
+
2
B
)
t
2
+
(
4
A
+
B
+
D
)
t
+
A
.
{\displaystyle 2t^{2}+2t+2=A(1+2t)^{2}+Bt(1+2t)+Dt=A(1+4t+4t^{2})+Bt+2Bt^{2}+Dt=(4A+2B)t^{2}+(4A+B+D)t+A.}
Prilyginę koeficientus prie vienodų laipsmių t , gauname sistemą lygčių pirmojo laipsnio atžvilgiu A, B, D:
{4A+2B=2,
{4A+B+D=2,
{A=2,
Iš čia A=2, B=-3, D=-3. Todėl,
2
t
2
+
2
t
+
2
t
(
1
+
2
t
)
2
=
2
t
−
3
1
+
2
t
−
3
(
1
+
2
t
)
2
,
{\displaystyle {\frac {2t^{2}+2t+2}{t(1+2t)^{2}}}={\frac {2}{t}}-{\frac {3}{1+2t}}-{\frac {3}{(1+2t)^{2}}},}
ir galutinai
∫
d
x
x
+
x
2
+
x
+
1
=
∫
[
2
t
−
3
1
+
2
t
−
3
(
1
+
2
t
)
2
]
d
t
=
2
∫
d
t
t
−
3
2
∫
d
(
1
+
2
t
)
1
+
2
t
−
3
2
∫
d
(
1
+
2
t
)
(
1
+
2
t
)
2
=
{\displaystyle \int {\frac {dx}{x+{\sqrt {x^{2}+x+1}}}}=\int [{\frac {2}{t}}-{\frac {3}{1+2t}}-{\frac {3}{(1+2t)^{2}}}]dt=2\int {\frac {dt}{t}}-{\frac {3}{2}}\int {\frac {d(1+2t)}{1+2t}}-{\frac {3}{2}}\int {\frac {d(1+2t)}{(1+2t)^{2}}}=}
=
2
ln
|
t
|
−
3
2
ln
|
1
+
2
t
|
+
3
2
(
1
+
2
t
)
+
C
=
{\displaystyle =2\ln |t|-{\frac {3}{2}}\ln |1+2t|+{\frac {3}{2(1+2t)}}+C=}
=
2
ln
|
x
+
x
2
+
x
+
1
|
−
3
2
ln
|
1
+
2
x
+
2
x
2
+
x
+
1
|
+
3
2
+
4
x
+
4
x
2
+
x
+
1
+
C
.
{\displaystyle =2\ln |x+{\sqrt {x^{2}+x+1}}|-{\frac {3}{2}}\ln |1+2x+2{\sqrt {x^{2}+x+1}}|+{\frac {3}{2+4x+4{\sqrt {x^{2}+x+1}}}}+C.}
Antrojo Oilerio keitinio pavyzdžiai
∫
d
x
1
−
6
x
−
9
x
2
;
{\displaystyle \int {\frac {dx}{\sqrt {1-6x-9x^{2}}}};}
1
−
6
x
−
9
x
2
=
u
x
+
1.
{\displaystyle {\sqrt {1-6x-9x^{2}}}=ux+1.}
1
−
6
x
−
9
x
2
=
u
2
x
2
+
2
u
x
+
1
;
{\displaystyle 1-6x-9x^{2}=u^{2}x^{2}+2ux+1;}
−
6
x
−
9
x
2
=
u
2
x
2
+
2
u
x
;
{\displaystyle -6x-9x^{2}=u^{2}x^{2}+2ux;}
−
6
−
9
x
=
u
2
x
+
2
u
.
{\displaystyle -6-9x=u^{2}x+2u.}
−
9
d
x
=
2
u
x
d
u
+
2
d
u
;
{\displaystyle -9dx=2uxdu+2du;}
d
x
=
−
2
(
u
x
+
1
)
9
+
u
2
d
u
.
{\displaystyle dx={\frac {-2(ux+1)}{9+u^{2}}}du.}
∫
d
x
1
−
6
x
−
9
x
2
=
−
2
∫
(
u
x
+
1
)
d
u
(
9
+
u
2
)
(
u
x
+
1
)
=
−
2
∫
d
u
9
+
u
2
=
−
2
3
arctan
u
3
+
C
=
{\displaystyle \int {\frac {dx}{\sqrt {1-6x-9x^{2}}}}=-2\int {\frac {(ux+1)du}{(9+u^{2})(ux+1)}}=-2\int {\frac {du}{9+u^{2}}}=-{\frac {2}{3}}\arctan {\frac {u}{3}}+C=}
=
−
2
3
arctan
1
−
6
x
−
9
x
2
−
1
3
x
+
C
.
{\displaystyle =-{\frac {2}{3}}\arctan {\frac {{\sqrt {1-6x-9x^{2}}}-1}{3x}}+C.}
Sprendimas normaliai . Čia a =-9, b =-6, c =1. Tada
x
=
2
c
t
−
b
a
−
t
2
=
2
1
t
−
(
−
6
)
−
9
−
t
2
=
2
t
+
6
−
9
−
t
2
.
{\displaystyle x={\frac {2{\sqrt {c}}t-b}{a-t^{2}}}={\frac {2{\sqrt {1}}t-(-6)}{-9-t^{2}}}={\frac {2t+6}{-9-t^{2}}}.}
1
−
6
x
−
9
x
2
=
x
t
+
1
=
2
t
+
6
−
9
−
t
2
t
+
1
=
2
t
2
+
6
t
−
9
−
t
2
+
1
=
2
t
2
+
6
t
−
9
−
t
2
−
9
−
t
2
=
t
2
+
6
t
−
9
−
9
−
t
2
.
{\displaystyle {\sqrt {1-6x-9x^{2}}}=xt+1={\frac {2t+6}{-9-t^{2}}}t+1={\frac {2t^{2}+6t}{-9-t^{2}}}+1={\frac {2t^{2}+6t-9-t^{2}}{-9-t^{2}}}={\frac {t^{2}+6t-9}{-9-t^{2}}}.}
d
x
=
2
c
t
2
−
2
b
t
+
2
a
c
(
a
−
t
2
)
2
d
t
=
2
1
t
2
−
2
(
−
6
)
t
+
2
(
−
9
)
1
(
−
9
−
t
2
)
2
d
t
=
2
t
2
+
12
t
−
18
(
−
9
−
t
2
)
2
d
t
=
2
(
t
2
+
6
t
−
9
)
(
−
9
−
t
2
)
2
d
t
.
{\displaystyle dx={\frac {2{\sqrt {c}}t^{2}-2bt+2a{\sqrt {c}}}{(a-t^{2})^{2}}}dt={\frac {2{\sqrt {1}}t^{2}-2(-6)t+2(-9){\sqrt {1}}}{(-9-t^{2})^{2}}}dt={\frac {2t^{2}+12t-18}{(-9-t^{2})^{2}}}dt={\frac {2(t^{2}+6t-9)}{(-9-t^{2})^{2}}}dt.}
∫
d
x
1
−
6
x
−
9
x
2
=
∫
2
(
t
2
+
6
t
−
9
)
(
−
9
−
t
2
)
2
d
t
t
2
+
6
t
−
9
−
9
−
t
2
=
∫
2
−
9
−
t
2
d
t
1
=
∫
2
d
t
−
9
−
t
2
=
−
2
∫
d
t
9
+
t
2
=
−
2
3
arctan
t
3
+
C
=
{\displaystyle \int {\frac {dx}{\sqrt {1-6x-9x^{2}}}}=\int {\frac {{\frac {2(t^{2}+6t-9)}{(-9-t^{2})^{2}}}dt}{\frac {t^{2}+6t-9}{-9-t^{2}}}}=\int {\frac {{\frac {2}{-9-t^{2}}}dt}{1}}=\int {\frac {2dt}{-9-t^{2}}}=-2\int {\frac {dt}{9+t^{2}}}=-{\frac {2}{3}}\arctan {\frac {t}{3}}+C=}
=
−
2
3
arctan
1
−
6
x
−
9
x
2
−
1
3
x
+
C
.
{\displaystyle =-{\frac {2}{3}}\arctan {\frac {{\sqrt {1-6x-9x^{2}}}-1}{3x}}+C.}
Apskaičiuoti
∫
d
x
(
1
+
x
)
1
+
x
−
x
2
.
{\displaystyle \int {\frac {dx}{(1+x){\sqrt {1+x-x^{2}}}}}.}
Sprendimas . Čia trinaris
1
+
x
−
x
2
{\displaystyle 1+x-x^{2}}
a <0, c >0, pasinaudojame keitiniu
1
+
x
−
x
2
=
t
x
−
1.
{\displaystyle {\sqrt {1+x-x^{2}}}=tx-1.}
1
+
x
−
x
2
=
t
2
x
2
−
2
t
x
+
1
{\displaystyle 1+x-x^{2}=t^{2}x^{2}-2tx+1}
x
−
x
2
=
t
2
x
2
−
2
t
x
{\displaystyle x-x^{2}=t^{2}x^{2}-2tx}
1
−
x
=
t
2
x
−
2
t
{\displaystyle 1-x=t^{2}x-2t}
1
+
2
t
=
t
2
x
+
x
{\displaystyle 1+2t=t^{2}x+x}
x
(
t
2
+
1
)
=
1
+
2
t
{\displaystyle x(t^{2}+1)=1+2t}
x
=
1
+
2
t
t
2
+
1
{\displaystyle x={\frac {1+2t}{t^{2}+1}}}
d
x
=
(
1
+
2
t
t
2
+
1
)
′
=
2
(
t
2
+
1
)
−
(
1
+
2
t
)
⋅
2
t
(
t
2
+
1
)
2
d
t
=
2
t
2
+
2
−
2
t
−
4
t
2
(
t
2
+
1
)
2
d
t
=
2
−
2
t
−
2
t
2
(
t
2
+
1
)
2
d
t
=
2
(
1
−
t
−
t
2
)
(
t
2
+
1
)
2
d
t
.
{\displaystyle dx=({\frac {1+2t}{t^{2}+1}})'={\frac {2(t^{2}+1)-(1+2t)\cdot 2t}{(t^{2}+1)^{2}}}dt={\frac {2t^{2}+2-2t-4t^{2}}{(t^{2}+1)^{2}}}dt={\frac {2-2t-2t^{2}}{(t^{2}+1)^{2}}}dt={\frac {2(1-t-t^{2})}{(t^{2}+1)^{2}}}dt.}
1
+
x
−
x
2
=
t
⋅
1
+
2
t
t
2
+
1
−
1
=
t
+
2
t
2
−
t
2
−
1
t
2
+
1
=
t
+
t
2
−
1
t
2
+
1
.
{\displaystyle {\sqrt {1+x-x^{2}}}=t\cdot {\frac {1+2t}{t^{2}+1}}-1={\frac {t+2t^{2}-t^{2}-1}{t^{2}+1}}={\frac {t+t^{2}-1}{t^{2}+1}}.}
1
+
x
−
x
2
=
t
x
−
1
;
1
+
x
−
x
2
+
1
=
t
x
;
t
=
1
+
x
−
x
2
+
1
x
.
{\displaystyle {\sqrt {1+x-x^{2}}}=tx-1;\;{\sqrt {1+x-x^{2}}}+1=tx;\;t={\frac {{\sqrt {1+x-x^{2}}}+1}{x}}.}
Tokiu budu,
∫
d
x
(
1
+
x
)
1
+
x
−
x
2
=
∫
2
(
1
−
t
−
t
2
)
(
t
2
+
1
)
2
d
t
(
1
+
1
+
2
t
t
2
+
1
)
t
+
t
2
−
1
t
2
+
1
=
∫
−
2
(
t
2
+
t
−
1
)
(
t
2
+
1
)
2
d
t
t
2
+
1
+
1
+
2
t
t
2
+
1
⋅
t
2
+
t
−
1
t
2
+
1
=
∫
−
2
d
t
1
+
t
2
+
2
t
+
1
=
{\displaystyle \int {\frac {dx}{(1+x){\sqrt {1+x-x^{2}}}}}=\int {\frac {{\frac {2(1-t-t^{2})}{(t^{2}+1)^{2}}}dt}{(1+{\frac {1+2t}{t^{2}+1}}){\frac {t+t^{2}-1}{t^{2}+1}}}}=\int {\frac {{\frac {-2(t^{2}+t-1)}{(t^{2}+1)^{2}}}dt}{{\frac {t^{2}+1+1+2t}{t^{2}+1}}\cdot {\frac {t^{2}+t-1}{t^{2}+1}}}}=\int {\frac {-2dt}{1+t^{2}+2t+1}}=}
=
∫
−
2
d
(
t
+
1
)
1
+
(
t
+
1
)
2
=
−
2
arctan
(
t
+
1
)
+
C
=
−
2
arctan
(
1
+
x
−
x
2
+
1
x
+
1
)
+
C
=
{\displaystyle =\int {\frac {-2d(t+1)}{1+(t+1)^{2}}}=-2\arctan(t+1)+C=-2\arctan({\frac {{\sqrt {1+x-x^{2}}}+1}{x}}+1)+C=}
=
−
2
arctan
1
+
x
−
x
2
+
1
+
x
x
+
C
.
{\displaystyle =-2\arctan {\frac {{\sqrt {1+x-x^{2}}}+1+x}{x}}+C.}
Reikia apskaičiuoti integralą
∫
(
1
−
1
+
x
+
x
2
)
2
x
2
1
+
x
+
x
2
d
x
.
{\displaystyle \int {\frac {(1-{\sqrt {1+x+x^{2}}})^{2}}{x^{2}{\sqrt {1+x+x^{2}}}}}dx.}
Sprendimas . Taikome II Oilerio keitinį
1
+
x
+
x
2
=
x
t
+
1
;
{\displaystyle {\sqrt {1+x+x^{2}}}=xt+1;}
1
+
x
+
x
2
=
x
2
t
2
+
2
x
t
+
1
;
{\displaystyle 1+x+x^{2}=x^{2}t^{2}+2xt+1;}
x
+
x
2
=
x
2
t
2
+
2
x
t
;
{\displaystyle x+x^{2}=x^{2}t^{2}+2xt;}
1
+
x
=
x
t
2
+
2
t
;
{\displaystyle 1+x=xt^{2}+2t;}
x
−
x
t
2
=
2
t
−
1
;
{\displaystyle x-xt^{2}=2t-1;}
x
(
1
−
t
2
)
=
2
t
−
1
;
{\displaystyle x(1-t^{2})=2t-1;}
x
=
2
t
−
1
1
−
t
2
.
{\displaystyle x={\frac {2t-1}{1-t^{2}}}.}
d
x
=
2
(
1
−
t
2
)
−
(
2
t
−
1
)
⋅
(
−
2
t
)
(
1
−
t
2
)
2
d
t
=
2
−
2
t
2
+
2
t
(
2
t
−
1
)
(
1
−
t
2
)
2
d
t
=
2
−
2
t
2
+
4
t
2
−
2
t
(
1
−
t
2
)
2
d
t
=
2
t
2
−
2
t
+
2
(
1
−
t
2
)
2
d
t
.
{\displaystyle dx={\frac {2(1-t^{2})-(2t-1)\cdot (-2t)}{(1-t^{2})^{2}}}dt={\frac {2-2t^{2}+2t(2t-1)}{(1-t^{2})^{2}}}dt={\frac {2-2t^{2}+4t^{2}-2t}{(1-t^{2})^{2}}}dt={\frac {2t^{2}-2t+2}{(1-t^{2})^{2}}}dt.}
1
+
x
+
x
2
=
2
t
−
1
1
−
t
2
⋅
t
+
1
=
2
t
2
−
t
+
(
1
−
t
2
)
1
−
t
2
=
t
2
−
t
+
1
1
−
t
2
.
{\displaystyle {\sqrt {1+x+x^{2}}}={\frac {2t-1}{1-t^{2}}}\cdot t+1={\frac {2t^{2}-t+(1-t^{2})}{1-t^{2}}}={\frac {t^{2}-t+1}{1-t^{2}}}.}
1
−
1
+
x
+
x
2
=
1
−
(
x
t
+
1
)
=
−
x
t
=
−
2
t
−
1
1
−
t
2
⋅
t
=
−
2
t
2
+
t
1
−
t
2
.
{\displaystyle 1-{\sqrt {1+x+x^{2}}}=1-(xt+1)=-xt=-{\frac {2t-1}{1-t^{2}}}\cdot t={\frac {-2t^{2}+t}{1-t^{2}}}.}
t
=
1
+
x
+
x
2
−
1
x
.
{\displaystyle t={\frac {{\sqrt {1+x+x^{2}}}-1}{x}}.}
Įstate gautas išraiškas į pradinį integralą, randame:
∫
(
1
−
1
+
x
+
x
2
)
2
x
2
1
+
x
+
x
2
d
x
=
∫
(
−
2
t
2
+
t
)
2
(
1
−
t
2
)
2
(
1
−
t
2
)
(
2
t
2
−
2
t
+
2
)
(
1
−
t
2
)
2
(
2
t
−
1
)
2
(
t
2
−
t
+
1
)
(
1
−
t
2
)
2
d
t
=
2
∫
(
−
2
t
2
+
t
)
2
(
2
t
−
1
)
2
(
1
−
t
2
)
d
t
=
2
∫
t
2
1
−
t
2
d
t
=
{\displaystyle \int {\frac {(1-{\sqrt {1+x+x^{2}}})^{2}}{x^{2}{\sqrt {1+x+x^{2}}}}}dx=\int {\frac {(-2t^{2}+t)^{2}(1-t^{2})^{2}(1-t^{2})(2t^{2}-2t+2)}{(1-t^{2})^{2}(2t-1)^{2}(t^{2}-t+1)(1-t^{2})^{2}}}dt=2\int {\frac {(-2t^{2}+t)^{2}}{(2t-1)^{2}(1-t^{2})}}dt=2\int {\frac {t^{2}}{1-t^{2}}}dt=}
=
2
∫
(
1
1
−
t
2
−
1
)
d
t
=
2
(
1
2
ln
|
1
+
t
1
−
t
|
−
t
)
+
C
=
{\displaystyle =2\int ({\frac {1}{1-t^{2}}}-1)dt=2({\frac {1}{2}}\ln |{\frac {1+t}{1-t}}|-t)+C=}
=
ln
|
1
+
1
+
x
+
x
2
−
1
x
1
−
1
+
x
+
x
2
−
1
x
|
−
2
(
1
+
x
+
x
2
−
1
)
x
+
C
=
ln
|
x
+
1
+
x
+
x
2
−
1
x
x
−
1
+
x
+
x
2
+
1
x
|
−
2
(
1
+
x
+
x
2
−
1
)
x
+
C
=
{\displaystyle =\ln |{\frac {1+{\frac {{\sqrt {1+x+x^{2}}}-1}{x}}}{1-{\frac {{\sqrt {1+x+x^{2}}}-1}{x}}}}|-{\frac {2({\sqrt {1+x+x^{2}}}-1)}{x}}+C=\ln |{\frac {\frac {x+{\sqrt {1+x+x^{2}}}-1}{x}}{\frac {x-{\sqrt {1+x+x^{2}}}+1}{x}}}|-{\frac {2({\sqrt {1+x+x^{2}}}-1)}{x}}+C=}
=
ln
|
x
+
1
+
x
+
x
2
−
1
x
−
1
+
x
+
x
2
+
1
|
−
2
(
1
+
x
+
x
2
−
1
)
x
+
C
=
{\displaystyle =\ln |{\frac {x+{\sqrt {1+x+x^{2}}}-1}{x-{\sqrt {1+x+x^{2}}}+1}}|-{\frac {2({\sqrt {1+x+x^{2}}}-1)}{x}}+C=}
=
ln
|
2
x
+
2
1
+
x
+
x
2
+
1
|
−
2
(
1
+
x
+
x
2
−
1
)
x
+
C
.
{\displaystyle =\ln |2x+2{\sqrt {1+x+x^{2}}}+1|-{\frac {2({\sqrt {1+x+x^{2}}}-1)}{x}}+C.}
Trečiojo Oilerio keitinio pavyzdžiai
∫
d
x
x
−
x
2
+
5
x
−
6
.
{\displaystyle \int {\frac {dx}{x{\sqrt {-x^{2}+5x-6}}}}.}
(
x
−
2
)
(
3
−
x
)
=
u
(
x
−
2
)
.
{\displaystyle {\sqrt {(x-2)(3-x)}}=u(x-2).}
(
x
−
2
)
,
{\displaystyle (x-2),}
(
x
−
2
)
(
3
−
x
)
=
u
2
(
x
−
2
)
2
;
{\displaystyle (x-2)(3-x)=u^{2}(x-2)^{2};}
3
−
x
=
u
2
(
x
−
2
)
;
{\displaystyle 3-x=u^{2}(x-2);}
−
x
u
2
−
x
=
−
2
u
2
−
3
;
{\displaystyle -xu^{2}-x=-2u^{2}-3;}
x
(
−
u
2
−
1
)
=
−
2
u
2
−
3
;
{\displaystyle x(-u^{2}-1)=-2u^{2}-3;}
x
=
2
u
2
+
3
u
2
+
1
;
u
=
3
−
x
x
−
2
{\displaystyle x={\frac {2u^{2}+3}{u^{2}+1}};\;u={\sqrt {\frac {3-x}{x-2}}}}
d
x
=
4
u
(
u
2
+
1
)
−
2
u
(
2
u
2
+
3
)
(
u
2
+
1
)
2
d
u
=
4
u
3
+
4
u
−
4
u
3
−
6
u
(
u
2
+
1
)
2
d
u
=
−
2
u
d
u
(
u
2
+
1
)
2
.
{\displaystyle dx={\frac {4u(u^{2}+1)-2u(2u^{2}+3)}{(u^{2}+1)^{2}}}du={\frac {4u^{3}+4u-4u^{3}-6u}{(u^{2}+1)^{2}}}du={\frac {-2u\;du}{(u^{2}+1)^{2}}}.}
(
x
−
2
)
(
3
−
x
)
=
u
(
x
−
2
)
=
u
(
2
u
2
+
3
u
2
+
1
−
2
)
=
2
u
3
+
3
u
−
2
u
(
u
2
+
1
)
u
2
+
1
=
2
u
3
+
3
u
−
2
u
3
−
2
u
u
2
+
1
=
u
u
2
+
1
.
{\displaystyle {\sqrt {(x-2)(3-x)}}=u(x-2)=u({\frac {2u^{2}+3}{u^{2}+1}}-2)={\frac {2u^{3}+3u-2u(u^{2}+1)}{u^{2}+1}}={\frac {2u^{3}+3u-2u^{3}-2u}{u^{2}+1}}={\frac {u}{u^{2}+1}}.}
∫
d
x
x
−
x
2
+
5
x
−
6
=
∫
−
2
u
d
u
(
u
2
+
1
)
2
2
u
2
+
3
u
2
+
1
⋅
u
u
2
+
1
=
−
2
∫
d
u
3
+
2
u
2
=
−
∫
d
u
3
2
+
u
2
=
{\displaystyle \int {\frac {dx}{x{\sqrt {-x^{2}+5x-6}}}}=\int {\frac {\frac {-2u\;du}{(u^{2}+1)^{2}}}{{\frac {2u^{2}+3}{u^{2}+1}}\cdot {\frac {u}{u^{2}+1}}}}=-2\int {\frac {du}{3+2u^{2}}}=-\int {\frac {du}{{\frac {3}{2}}+u^{2}}}=}
=
−
1
3
2
arctan
(
u
3
2
)
+
C
=
−
2
3
arctan
(
2
3
u
)
+
C
=
−
2
3
arctan
(
2
3
⋅
3
−
x
x
−
2
)
+
C
.
{\displaystyle =-{\frac {1}{\sqrt {\frac {3}{2}}}}\arctan({\frac {u}{\sqrt {\frac {3}{2}}}})+C=-{\sqrt {\frac {2}{3}}}\arctan({\sqrt {\frac {2}{3}}}u)+C=-{\sqrt {\frac {2}{3}}}\arctan({\sqrt {\frac {2}{3}}}\cdot {\sqrt {\frac {3-x}{x-2}}})+C.}
∫
x
d
x
(
7
x
−
10
−
x
2
)
3
,
{\displaystyle \int {\frac {x\;dx}{\sqrt {(7x-10-x^{2})^{3}}}},}
a
<
0
,
{\displaystyle a<0,}
c
<
0
,
{\displaystyle c<0,}
7
x
−
10
−
x
2
{\displaystyle 7x-10-x^{2}}
x
1
=
2
{\displaystyle x_{1}=2}
x
2
=
5
{\displaystyle x_{2}=5}
a
=
−
1.
{\displaystyle a=-1.}
7
x
−
10
−
x
2
=
−
1
(
x
−
2
)
(
x
−
5
)
=
(
x
−
2
)
(
5
−
x
)
=
(
x
−
2
)
t
.
{\displaystyle {\sqrt {7x-10-x^{2}}}={\sqrt {-1(x-2)(x-5)}}={\sqrt {(x-2)(5-x)}}=(x-2)t.}
5
−
x
=
(
x
−
2
)
t
2
;
{\displaystyle 5-x=(x-2)t^{2};}
−
x
t
2
−
x
=
−
2
t
2
−
5
;
{\displaystyle -xt^{2}-x=-2t^{2}-5;}
x
=
5
+
2
t
2
1
+
t
2
;
{\displaystyle x={\frac {5+2t^{2}}{1+t^{2}}};}
d
x
=
(
5
+
2
t
2
1
+
t
2
)
′
=
4
t
(
1
+
t
2
)
−
(
5
+
2
t
2
)
2
t
(
1
+
t
2
)
2
d
t
=
4
t
+
4
t
3
−
10
t
−
4
t
3
(
1
+
t
2
)
2
d
t
=
−
6
t
(
1
+
t
2
)
2
d
t
.
{\displaystyle dx=({\frac {5+2t^{2}}{1+t^{2}}})'={\frac {4t(1+t^{2})-(5+2t^{2})2t}{(1+t^{2})^{2}}}dt={\frac {4t+4t^{3}-10t-4t^{3}}{(1+t^{2})^{2}}}dt={\frac {-6t}{(1+t^{2})^{2}}}dt.}
7
x
−
10
−
x
2
=
(
x
−
2
)
t
=
(
5
+
2
t
2
1
+
t
2
−
2
)
t
=
5
+
2
t
2
−
2
−
2
t
2
1
+
t
2
t
=
3
t
1
+
t
2
.
{\displaystyle {\sqrt {7x-10-x^{2}}}=(x-2)t=({\frac {5+2t^{2}}{1+t^{2}}}-2)t={\frac {5+2t^{2}-2-2t^{2}}{1+t^{2}}}t={\frac {3t}{1+t^{2}}}.}
t
=
(
x
−
2
)
(
5
−
x
)
(
x
−
2
)
=
(
5
−
x
)
(
x
−
2
)
.
{\displaystyle t={\frac {\sqrt {(x-2)(5-x)}}{(x-2)}}={\sqrt {\frac {(5-x)}{(x-2)}}}.}
∫
x
d
x
(
7
x
−
10
−
x
2
)
3
=
∫
5
+
2
t
2
1
+
t
2
−
6
t
(
1
+
t
2
)
2
d
t
(
3
t
1
+
t
2
)
3
=
∫
−
6
t
(
5
+
2
t
2
)
(
1
+
t
2
)
3
(
3
t
1
+
t
2
)
3
d
t
=
∫
−
6
t
(
5
+
2
t
2
)
27
t
3
d
t
=
∫
−
6
(
5
+
2
t
2
)
27
t
2
d
t
=
{\displaystyle \int {\frac {x\;dx}{\sqrt {(7x-10-x^{2})^{3}}}}=\int {\frac {{\frac {5+2t^{2}}{1+t^{2}}}{\frac {-6t}{(1+t^{2})^{2}}}dt}{({\frac {3t}{1+t^{2}}})^{3}}}=\int {\frac {\frac {-6t(5+2t^{2})}{(1+t^{2})^{3}}}{({\frac {3t}{1+t^{2}}})^{3}}}dt=\int {\frac {-6t(5+2t^{2})}{27t^{3}}}dt=\int {\frac {-6(5+2t^{2})}{27t^{2}}}dt=}
=
∫
(
−
30
27
t
2
−
12
t
2
27
t
2
)
d
t
=
1
27
∫
(
−
30
t
2
−
12
)
d
t
=
1
27
(
30
t
−
12
t
)
+
C
=
1
27
(
30
(
5
−
x
)
(
x
−
2
)
−
12
(
5
−
x
)
(
x
−
2
)
)
+
C
=
{\displaystyle =\int (-{\frac {30}{27t^{2}}}-{\frac {12t^{2}}{27t^{2}}})dt={\frac {1}{27}}\int (-{\frac {30}{t^{2}}}-12)dt={\frac {1}{27}}({\frac {30}{t}}-12t)+C={\frac {1}{27}}({\frac {30}{\sqrt {\frac {(5-x)}{(x-2)}}}}-12{\sqrt {\frac {(5-x)}{(x-2)}}})+C=}
=
30
x
−
2
27
5
−
x
−
12
27
(
5
−
x
)
(
x
−
2
)
+
C
.
{\displaystyle ={\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {(5-x)}{(x-2)}}}+C.}
Reikia apskaičiuoti integralą
∫
d
x
x
2
+
3
x
−
4
.
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3x-4}}}.}
Sprendimas . Kadangi
x
2
+
3
x
−
4
=
(
x
+
4
)
(
x
−
1
)
,
{\displaystyle x^{2}+3x-4=(x+4)(x-1),}
(
x
+
4
)
(
x
−
1
)
=
(
x
+
4
)
t
;
{\displaystyle {\sqrt {(x+4)(x-1)}}=(x+4)t;}
(
x
+
4
)
(
x
−
1
)
=
(
x
+
4
)
2
t
2
{\displaystyle (x+4)(x-1)=(x+4)^{2}t^{2}}
x
−
1
=
(
x
+
4
)
t
2
,
{\displaystyle x-1=(x+4)t^{2},}
x
−
1
=
x
t
2
+
4
t
2
,
{\displaystyle x-1=xt^{2}+4t^{2},}
x
−
x
t
2
=
4
t
2
+
1
,
{\displaystyle x-xt^{2}=4t^{2}+1,}
x
=
1
+
4
t
2
1
−
t
2
.
{\displaystyle x={\frac {1+4t^{2}}{1-t^{2}}}.}
d
x
=
(
1
+
4
t
2
)
′
⋅
(
1
−
t
2
)
−
(
1
+
4
t
2
)
⋅
(
1
−
t
2
)
′
(
1
−
t
2
)
2
=
8
t
⋅
(
1
−
t
2
)
−
(
1
+
4
t
2
)
⋅
(
−
2
t
)
(
1
−
t
2
)
2
d
t
=
{\displaystyle dx={\frac {(1+4t^{2})'\cdot (1-t^{2})-(1+4t^{2})\cdot (1-t^{2})'}{(1-t^{2})^{2}}}={\frac {8t\cdot (1-t^{2})-(1+4t^{2})\cdot (-2t)}{(1-t^{2})^{2}}}dt=}
=
8
t
−
8
t
3
+
2
t
+
8
t
3
(
1
−
t
2
)
2
d
t
=
10
t
(
1
−
t
2
)
2
d
t
.
{\displaystyle ={\frac {8t-8t^{3}+2t+8t^{3}}{(1-t^{2})^{2}}}dt={\frac {10t}{(1-t^{2})^{2}}}dt.}
(
x
+
4
)
(
x
−
1
)
=
(
x
+
4
)
t
=
(
1
+
4
t
2
1
−
t
2
+
4
)
t
=
1
+
4
t
2
+
4
−
4
t
2
1
−
t
2
⋅
t
=
5
t
1
−
t
2
.
{\displaystyle {\sqrt {(x+4)(x-1)}}=(x+4)t=({\frac {1+4t^{2}}{1-t^{2}}}+4)t={\frac {1+4t^{2}+4-4t^{2}}{1-t^{2}}}\cdot t={\frac {5t}{1-t^{2}}}.}
t
=
(
x
+
4
)
(
x
−
1
)
(
x
+
4
)
=
(
x
+
4
)
(
x
−
1
)
(
x
+
4
)
2
=
x
−
1
x
+
4
.
{\displaystyle t={\frac {\sqrt {(x+4)(x-1)}}{(x+4)}}={\sqrt {\frac {(x+4)(x-1)}{(x+4)^{2}}}}={\sqrt {\frac {x-1}{x+4}}}.}
Grįžtant prie pradinio integralo, gauname:
∫
d
x
x
2
+
3
x
−
4
=
∫
10
t
(
1
−
t
2
)
(
1
−
t
2
)
2
5
t
d
t
=
∫
2
1
−
t
2
d
t
=
ln
|
1
+
t
1
−
t
|
+
C
=
ln
|
1
+
x
−
1
x
+
4
1
−
x
−
1
x
+
4
|
+
C
=
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3x-4}}}=\int {\frac {10t(1-t^{2})}{(1-t^{2})^{2}5t}}dt=\int {\frac {2}{1-t^{2}}}dt=\ln |{\frac {1+t}{1-t}}|+C=\ln |{\frac {1+{\sqrt {\frac {x-1}{x+4}}}}{1-{\sqrt {\frac {x-1}{x+4}}}}}|+C=}
=
ln
|
x
+
4
+
x
−
1
x
+
4
−
x
−
1
|
+
C
.
{\displaystyle =\ln |{\frac {{\sqrt {x+4}}+{\sqrt {x-1}}}{{\sqrt {x+4}}-{\sqrt {x-1}}}}|+C.}
Diferencialinių binomų integravimas
keisti
Integralas
∫
x
m
(
a
+
b
x
n
)
p
d
x
,
{\displaystyle \int x^{m}(a+bx^{n})^{p}dx,}
m, n, p - racionalieji skaičiai, vadinamas integralu su binominiu diferencialu.
Šį integralą elementariosiomis funkcijomis įmanoma išreikšti tik trimis atvejais:
I. p - sveikasis skaičius. Jei
p
>
0
,
{\displaystyle p>0,}
Niutono binomo formulę . Jei
p
<
0
,
{\displaystyle p<0,}
x
=
t
k
,
{\displaystyle x=t^{k},}
k - bendras trupmenų m ir n vardiklis. Pavyzdžiui, trupmenų
1
4
{\displaystyle {\tfrac {1}{4}}}
2
3
{\displaystyle {\tfrac {2}{3}}}
⋅
{\displaystyle \cdot }
II.
m
+
1
n
{\displaystyle {\frac {m+1}{n}}}
a
+
b
x
n
=
t
α
,
{\displaystyle a+bx^{n}=t^{\alpha },}
α
{\displaystyle \alpha }
p vardiklis.
III.
m
+
1
n
+
p
{\displaystyle {\frac {m+1}{n}}+p}
a
+
b
x
n
=
t
α
x
n
,
{\displaystyle a+bx^{n}=t^{\alpha }x^{n},}
α
{\displaystyle \alpha }
p vardiklis. Pavyzdžiai
∫
x
1
3
(
2
+
x
)
2
d
x
=
∫
x
1
3
(
4
+
4
x
+
x
)
d
x
=
∫
(
4
x
1
3
+
4
x
5
6
+
x
4
3
)
d
x
=
3
x
4
3
+
24
11
x
11
6
+
3
7
x
7
3
+
C
,
{\displaystyle \int x^{\frac {1}{3}}(2+{\sqrt {x}})^{2}dx=\int x^{\frac {1}{3}}(4+4{\sqrt {x}}+x)dx=\int (4x^{\frac {1}{3}}+4x^{\frac {5}{6}}+x^{\frac {4}{3}})dx=3x^{\frac {4}{3}}+{\frac {24}{11}}x^{\frac {11}{6}}+{\frac {3}{7}}x^{\frac {7}{3}}+C,}
kur
p
=
2
{\displaystyle p=2}
∫
1
+
x
1
3
x
2
3
d
x
=
∫
x
−
2
3
(
1
+
x
1
3
)
1
2
d
x
=
∫
t
2
6
t
d
t
=
6
∫
t
2
d
t
=
2
t
3
+
C
=
2
(
1
+
x
1
3
)
3
2
+
C
,
{\displaystyle \int {\frac {\sqrt {1+x^{\frac {1}{3}}}}{x^{\frac {2}{3}}}}dx=\int x^{-{\frac {2}{3}}}(1+x^{\frac {1}{3}})^{\frac {1}{2}}dx=\int {\sqrt {t^{2}}}6t\;dt=6\int t^{2}\;dt=2t^{3}+C=2(1+x^{\frac {1}{3}})^{\frac {3}{2}}+C,}
kur
m
=
−
2
3
;
n
=
1
3
;
p
=
1
2
;
m
+
1
n
=
1.
{\displaystyle m=-{\frac {2}{3}};\;n={\frac {1}{3}};\;p={\frac {1}{2}};\;{\frac {m+1}{n}}=1.}
1
+
x
1
3
=
t
2
;
1
3
x
−
2
3
d
x
=
2
t
d
t
.
{\displaystyle 1+x^{\frac {1}{3}}=t^{2};\;{\frac {1}{3}}x^{-{\frac {2}{3}}}dx=2t\;dt.}
∫
x
−
11
(
1
+
x
4
)
−
1
2
d
x
=
−
1
2
∫
(
t
2
−
1
)
11
4
(
t
2
t
2
−
1
)
−
1
2
t
d
t
(
t
2
−
1
)
5
4
=
−
1
2
∫
(
t
2
−
1
)
11
4
+
1
2
−
5
4
t
t
2
2
d
t
=
{\displaystyle \int x^{-11}(1+x^{4})^{-{\frac {1}{2}}}dx=-{\frac {1}{2}}\int (t^{2}-1)^{\frac {11}{4}}({\frac {t^{2}}{t^{2}-1}})^{-{\frac {1}{2}}}{\frac {t\;dt}{(t^{2}-1)^{\frac {5}{4}}}}=-{\frac {1}{2}}\int (t^{2}-1)^{{\frac {11}{4}}+{\frac {1}{2}}-{\frac {5}{4}}}{\frac {t}{t^{\frac {2}{2}}}}dt=}
=
−
1
2
∫
(
t
2
−
1
)
2
d
t
=
−
t
5
10
+
t
3
3
−
t
2
+
C
=
−
(
1
+
x
4
)
5
2
10
x
2
+
(
1
+
x
4
)
3
2
3
x
2
−
(
1
+
x
4
)
1
2
2
x
2
+
C
,
{\displaystyle =-{\frac {1}{2}}\int (t^{2}-1)^{2}dt=-{\frac {t^{5}}{10}}+{\frac {t^{3}}{3}}-{\frac {t}{2}}+C=-{\frac {(1+x^{4})^{\frac {5}{2}}}{10x^{2}}}+{\frac {(1+x^{4})^{\frac {3}{2}}}{3x^{2}}}-{\frac {(1+x^{4})^{\frac {1}{2}}}{2x^{2}}}+C,}
m
=
−
11
;
{\displaystyle m=-11;}
n
=
4
;
{\displaystyle n=4;}
p
=
−
1
2
;
m
+
1
n
+
p
=
−
3.
{\displaystyle p=-{\frac {1}{2}};\;{\frac {m+1}{n}}+p=-3.}
1
+
x
4
=
x
4
t
2
;
{\displaystyle 1+x^{4}=x^{4}t^{2};}
x
=
1
(
t
2
−
1
)
1
4
;
d
x
=
−
t
d
t
2
(
t
2
−
1
)
5
4
;
t
=
1
+
x
4
x
2
.
{\displaystyle x={\frac {1}{(t^{2}-1)^{\frac {1}{4}}}};\;dx=-{\frac {t\;dt}{2(t^{2}-1)^{\frac {5}{4}}}};\;t={\frac {\sqrt {1+x^{4}}}{x^{2}}}.}
∫
x
3
d
x
1
+
2
x
2
=
∫
x
3
(
1
+
2
x
2
)
−
1
2
d
x
=
1
2
∫
(
u
2
−
1
2
)
3
2
u
−
1
(
u
2
−
1
2
)
−
1
2
u
d
u
=
1
2
∫
u
2
−
1
2
d
u
=
{\displaystyle \int {\frac {x^{3}\;dx}{\sqrt {1+2x^{2}}}}=\int x^{3}(1+2x^{2})^{-{\frac {1}{2}}}dx={\frac {1}{2}}\int ({\frac {u^{2}-1}{2}})^{\frac {3}{2}}u^{-1}({\frac {u^{2}-1}{2}})^{-{\frac {1}{2}}}u\;du={\frac {1}{2}}\int {\frac {u^{2}-1}{2}}du=}
=
1
4
(
u
3
3
−
u
)
+
C
=
(
1
+
2
x
2
)
3
2
12
−
1
+
2
x
2
4
+
C
,
{\displaystyle ={\frac {1}{4}}({\frac {u^{3}}{3}}-u)+C={\frac {(1+2x^{2})^{\frac {3}{2}}}{12}}-{\frac {\sqrt {1+2x^{2}}}{4}}+C,}
m
+
1
n
=
3
+
1
2
=
2
;
{\displaystyle {\frac {m+1}{n}}={\frac {3+1}{2}}=2;}
1
+
2
x
2
=
u
2
;
{\displaystyle 1+2x^{2}=u^{2};}
x
=
(
u
2
−
1
2
)
1
2
;
d
x
=
1
2
(
u
2
−
1
2
)
−
1
2
u
d
u
.
{\displaystyle x=({\frac {u^{2}-1}{2}})^{\frac {1}{2}};\;dx={\frac {1}{2}}({\frac {u^{2}-1}{2}})^{-{\frac {1}{2}}}u\;du.}
∫
x
−
1
3
(
1
−
x
2
3
)
−
1
2
d
x
=
−
3
∫
t
−
1
⋅
t
d
t
=
−
3
∫
d
t
=
−
3
t
+
C
=
−
3
1
−
x
2
3
+
C
,
{\displaystyle \int x^{-{\frac {1}{3}}}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}dx=-3\int t^{-1}\cdot t\;dt=-3\int dt=-3t+C=-3{\sqrt {1-x^{\frac {2}{3}}}}+C,}
kur
m
=
−
1
3
;
{\displaystyle m=-{\frac {1}{3}};}
n
=
2
3
;
p
=
−
1
2
;
m
+
1
n
=
1
;
1
−
x
2
3
=
t
2
;
−
2
3
x
−
1
3
d
x
=
2
t
d
t
;
d
x
=
−
3
x
1
3
t
d
t
;
x
2
3
=
1
−
t
2
.
{\displaystyle n={\frac {2}{3}};\;p=-{\frac {1}{2}};\;{\frac {m+1}{n}}=1;\;1-x^{\frac {2}{3}}=t^{2};\;-{\frac {2}{3}}x^{-{\frac {1}{3}}}dx=2t\;dt;\;dx=-3x^{\frac {1}{3}}tdt;\;x^{\frac {2}{3}}=1-t^{2}.}
∫
x
1
3
(
1
−
x
2
3
)
−
1
2
d
x
=
−
3
∫
x
1
3
⋅
(
t
2
)
−
1
2
⋅
x
1
3
⋅
t
d
t
=
−
3
∫
x
2
3
⋅
t
−
1
⋅
t
d
t
=
{\displaystyle \int x^{\frac {1}{3}}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}dx=-3\int x^{\frac {1}{3}}\cdot (t^{2})^{-{\frac {1}{2}}}\cdot x^{\frac {1}{3}}\cdot t\;dt=-3\int x^{\frac {2}{3}}\cdot t^{-1}\cdot t\;dt=}
=
−
3
∫
(
1
−
t
2
)
t
t
d
t
=
−
3
∫
(
1
−
t
2
)
d
t
=
−
3
(
t
−
t
3
3
)
+
C
=
−
3
(
1
−
x
2
3
−
(
1
−
x
2
3
)
3
3
)
+
C
=
{\displaystyle =-3\int (1-t^{2}){\frac {t}{t}}\;dt=-3\int (1-t^{2})\;dt=-3(t-{\frac {t^{3}}{3}})+C=-3({\sqrt {1-x^{\frac {2}{3}}}}-{\frac {\sqrt {(1-x^{\frac {2}{3}})^{3}}}{3}})+C=}
=
−
3
3
(
1
−
x
2
3
)
1
2
−
(
1
−
x
2
3
)
3
2
3
+
C
=
−
3
(
1
−
x
2
3
)
1
2
+
(
1
−
x
2
3
)
3
2
+
C
=
{\displaystyle =-3{\frac {3(1-x^{\frac {2}{3}})^{\frac {1}{2}}-(1-x^{\frac {2}{3}})^{\frac {3}{2}}}{3}}+C=-3(1-x^{\frac {2}{3}})^{\frac {1}{2}}+(1-x^{\frac {2}{3}})^{\frac {3}{2}}+C=}
=
−
3
1
−
x
2
3
+
1
−
3
x
2
3
+
3
x
4
3
−
x
2
+
C
,
{\displaystyle =-3{\sqrt {1-x^{\frac {2}{3}}}}+{\sqrt {1-3x^{\frac {2}{3}}+3x^{\frac {4}{3}}-x^{2}}}+C,}
kur
m
=
1
3
;
{\displaystyle m={\frac {1}{3}};}
n
=
2
3
;
p
=
−
1
2
;
m
+
1
n
=
2
;
1
−
x
2
3
=
t
2
;
−
2
3
x
−
1
3
d
x
=
2
t
d
t
;
d
x
=
−
3
x
1
3
t
d
t
;
x
2
3
=
1
−
t
2
.
{\displaystyle n={\frac {2}{3}};\;p=-{\frac {1}{2}};\;{\frac {m+1}{n}}=2;\;1-x^{\frac {2}{3}}=t^{2};\;-{\frac {2}{3}}x^{-{\frac {1}{3}}}dx=2t\;dt;\;dx=-3x^{\frac {1}{3}}tdt;\;x^{\frac {2}{3}}=1-t^{2}.}
∫
d
x
x
2
+
3
.
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3}}}.}
m
+
1
n
+
p
=
0
+
1
2
−
1
2
=
0
{\displaystyle {\frac {m+1}{n}}+p={\frac {0+1}{2}}-{\frac {1}{2}}=0}
p
=
−
1
2
{\displaystyle p=-{\frac {1}{2}}}
a
+
b
x
n
=
t
α
x
n
,
{\displaystyle a+bx^{n}=t^{\alpha }x^{n},}
α
{\displaystyle \alpha }
p vardiklis. Taigi
a
+
b
x
n
=
t
α
x
n
{\displaystyle a+bx^{n}=t^{\alpha }x^{n}}
x
2
+
3
=
t
2
x
2
{\displaystyle x^{2}+3=t^{2}x^{2}}
3
=
t
2
x
2
−
x
2
{\displaystyle 3=t^{2}x^{2}-x^{2}}
x
2
=
3
t
2
−
1
=
3
(
t
2
−
1
)
−
1
{\displaystyle x^{2}={\frac {3}{t^{2}-1}}=3(t^{2}-1)^{-1}}
x
=
3
⋅
(
t
2
−
1
)
−
1
2
{\displaystyle x={\sqrt {3}}\cdot (t^{2}-1)^{-{\frac {1}{2}}}}
d
x
=
−
3
2
⋅
(
t
2
−
1
)
−
3
2
⋅
2
t
d
t
=
−
3
t
(
t
2
−
1
)
3
d
t
.
{\displaystyle dx=-{\frac {\sqrt {3}}{2}}\cdot (t^{2}-1)^{-{\frac {3}{2}}}\cdot 2t\;dt=-{\frac {{\sqrt {3}}t}{\sqrt {(t^{2}-1)^{3}}}}dt.}
∫
d
x
x
2
+
3
=
−
3
∫
t
d
t
(
t
2
−
1
)
3
⋅
3
t
2
−
1
+
3
=
−
3
∫
t
d
t
(
t
2
−
1
)
3
⋅
3
1
+
t
2
−
1
t
2
−
1
=
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3}}}=-{\sqrt {3}}\int {\frac {t\;dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {{\frac {3}{t^{2}-1}}+3}}}}=-{\sqrt {3}}\int {\frac {t\;dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {3}}{\sqrt {\frac {1+t^{2}-1}{t^{2}-1}}}}}=}
=
−
∫
t
d
t
(
t
2
−
1
)
3
⋅
t
2
t
2
−
1
=
−
∫
t
d
t
(
t
2
−
1
)
3
⋅
t
1
t
2
−
1
=
−
∫
d
t
(
t
2
−
1
)
3
⋅
1
t
2
−
1
=
{\displaystyle =-\int {\frac {t\;dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {\frac {t^{2}}{t^{2}-1}}}}}=-\int {\frac {t\;dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot t{\sqrt {\frac {1}{t^{2}-1}}}}}=-\int {\frac {dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {\frac {1}{t^{2}-1}}}}}=}
=
−
∫
d
t
(
t
2
−
1
)
3
t
2
−
1
=
−
∫
d
t
(
t
2
−
1
)
2
=
−
∫
d
t
t
2
−
1
=
∫
d
t
1
−
t
2
.
{\displaystyle =-\int {\frac {dt}{\sqrt {\frac {(t^{2}-1)^{3}}{t^{2}-1}}}}=-\int {\frac {dt}{\sqrt {(t^{2}-1)^{2}}}}=-\int {\frac {dt}{t^{2}-1}}=\int {\frac {dt}{1-t^{2}}}.}
Iš interentinio integratoriaus:
∫
d
t
1
−
t
2
=
1
2
(
ln
(
t
+
1
)
−
ln
(
1
−
t
)
)
+
C
=
tanh
−
1
t
+
C
=
artanh
t
+
C
.
{\displaystyle \int {\frac {dt}{1-t^{2}}}={\frac {1}{2}}(\ln(t+1)-\ln(1-t))+C=\tanh ^{-1}t+C={\text{artanh}}\;t+C.}
Kur
t
=
x
2
+
3
x
2
.
{\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}}.}
t
=
x
2
+
3
x
2
,
{\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}},}
Galima taip integruot:
−
∫
d
t
t
2
−
1
=
−
∫
d
t
(
t
−
1
)
(
t
+
1
)
;
{\displaystyle -\int {\frac {dt}{t^{2}-1}}=-\int {\frac {dt}{(t-1)(t+1)}};}
toliau integruojama kaip racionali funkcija.
−
1
(
t
−
1
)
(
t
+
1
)
=
A
t
−
1
+
B
t
+
1
;
{\displaystyle {\frac {-1}{(t-1)(t+1)}}={\frac {A}{t-1}}+{\frac {B}{t+1}};}
Abi lygties puses padauginame iš (t-1)(t+1). Tada
−
1
=
A
(
t
+
1
)
+
B
(
t
−
1
)
,
{\displaystyle -1=A(t+1)+B(t-1),}
−
1
=
A
t
+
A
+
B
t
−
B
,
{\displaystyle -1=At+A+Bt-B,}
−
1
=
(
A
+
B
)
t
+
A
−
B
;
{\displaystyle -1=(A+B)t+A-B;}
iš čia turime sistemą:
A+B=0,
A-B=-1.
Tada iš antros lygties A=B-1. Įstačius šia A reikšmę į pirmą lygtį, gauname
B-1+B=0,
2B=1,
B=1/2.
Tada A=-B=-1/2.
Tokiu budu gauname, kad
−
1
(
t
−
1
)
(
t
+
1
)
=
−
1
/
2
t
−
1
+
1
/
2
t
+
1
.
{\displaystyle {\frac {-1}{(t-1)(t+1)}}={\frac {-1/2}{t-1}}+{\frac {1/2}{t+1}}.}
Integruodami gauname:
−
∫
d
t
t
2
−
1
=
−
∫
d
t
(
t
−
1
)
(
t
+
1
)
=
∫
(
−
1
/
2
t
−
1
+
1
/
2
t
+
1
)
d
t
=
1
2
(
−
ln
(
t
−
1
)
+
ln
(
t
+
1
)
)
+
C
.
{\displaystyle -\int {\frac {dt}{t^{2}-1}}=-\int {\frac {dt}{(t-1)(t+1)}}=\int ({\frac {-1/2}{t-1}}+{\frac {1/2}{t+1}})dt={\frac {1}{2}}(-\ln(t-1)+\ln(t+1))+C.}
Kur
t
=
x
2
+
3
x
2
.
{\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}}.}
1. Jei įstatysime x=2, tai gausime,
t
=
x
2
+
3
x
2
=
2
2
+
3
2
2
=
7
4
=
1.75
=
1.32287565553.
{\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}}={\sqrt {2^{2}+3 \over 2^{2}}}={\sqrt {7 \over 4}}={\sqrt {1.75}}=1.32287565553.}
1
2
(
−
ln
(
t
−
1
)
+
ln
(
t
+
1
)
)
=
1
2
(
−
ln
(
1.75
−
1
)
+
ln
(
1.75
+
1
)
)
=
0.986646961
=
{\displaystyle {\frac {1}{2}}(-\ln(t-1)+\ln(t+1))={\frac {1}{2}}(-\ln({\sqrt {1.75}}-1)+\ln({\sqrt {1.75}}+1))=0.986646961=}
=0.98664696104483410110205523811797.
Kai x=1, tai
t
=
x
2
+
3
x
2
=
1
2
+
3
1
2
=
4
1
=
4
=
2.
{\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}}={\sqrt {1^{2}+3 \over 1^{2}}}={\sqrt {4 \over 1}}={\sqrt {4}}=2.}
1
2
(
−
ln
(
t
−
1
)
+
ln
(
t
+
1
)
)
=
1
2
(
−
ln
(
2
−
1
)
+
ln
(
2
+
1
)
)
=
ln
3
2
=
0.549306144334
=
{\displaystyle {\frac {1}{2}}(-\ln(t-1)+\ln(t+1))={\frac {1}{2}}(-\ln(2-1)+\ln(2+1))={\frac {\ln 3}{2}}=0.549306144334=}
=0.54930614433405484569762261846126. Bet tokia funkcija integruojama lengviau kitaip (ne per diferencialinius binomus; integruojant keičiant kintamąjį) ir yra jinai integralų lentelėje
∫
d
x
x
2
±
a
2
=
ln
|
x
+
x
2
±
a
2
|
+
C
.
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}\pm a^{2}}}}=\ln |x+{\sqrt {x^{2}\pm a^{2}}}|+C.}
2. Tada, kai x=2, tai
ln
|
x
+
x
2
+
3
|
=
ln
|
2
+
2
2
+
3
|
=
ln
|
2
+
7
|
=
1.5359531
=
{\displaystyle \ln |x+{\sqrt {x^{2}+3}}|=\ln |2+{\sqrt {2^{2}+3}}|=\ln |2+{\sqrt {7}}|=1.5359531=}
=1.5359531053788889467996778565792.
O kai x=1, tai
ln
|
x
+
x
2
+
3
|
=
ln
|
1
+
1
2
+
3
|
=
ln
|
1
+
4
|
=
ln
|
3
|
=
1.0986122886681
=
{\displaystyle \ln |x+{\sqrt {x^{2}+3}}|=\ln |1+{\sqrt {1^{2}+3}}|=\ln |1+{\sqrt {4}}|=\ln |3|=1.0986122886681=}
=1.0986122886681096913952452369225. Tada, kai x kinta nuo 1 iki 2, tai pirmu atveju integruojant gauname:
0.98664696104483410110205523811797 - 0.54930614433405484569762261846126 = 0.43734081671077925540443261965671.
Antru atveju, kai x kinta nuo 1 iki 2 integruojant gautume:
1.5359531053788889467996778565792 - 1.0986122886681096913952452369225 = 0.4373408167107792554044326196567.
Abiais būdais integruojant gavome tą patį atsakymą. Toks Free Pascal kodas: var a:longint; c,d:real;
begin
for a:=1 to 100000000 do
d:=d+0.00000002/sqrt(sqr(a*0.00000002)+3);
for a:=1 to 100000000 do
c:=c+0.00000001/sqrt(sqr(a*0.00000001)+3);
writeln(d);
writeln(c);
writeln(d-c);
readln;
end.
Duoda rezultatus:
9.8664695905111544E-001
5.4930614394743249E-001
4.3734081510368294E-001
po 4 sekundžių su 4.16 GHz dažniu veikiančiu procesorium (per pirmus du paleidimus duoda šituos rezultatus po 18 sekundžių; bet jeigu iškart exe failą (diferencialiniaibinomai.exe) paleist [kurį sukuria visada Free Pascal programa] iš "C:\FPC\3.2.0\bin\i386-win32", tai rezultatai gaunami po 4 sekundžių ir taip yra su visais Free Pascal skaičiavimais, kad per exe failą greičiau skaičiuoja [iš pirmo karto]). Matome, kad rezultatai tokie patys kaip skaičiuojant/integruojant pirmu atveju. Toks Free Pascal kodas: var a:longint; c,d:real;
begin
for a:=1 to 1000000000 do
d:=d+0.000000002/sqrt(sqr(a*0.000000002)+3);
writeln(d);
readln;
end.
duoda rezultatą "9.8664696084608883E-001" (tai reiškia
9.8664696084608883
⋅
10
−
1
{\displaystyle 9.8664696084608883\cdot 10^{-1}}
Kodas apskaičiuoja plotą, po funkciją
f
(
x
)
=
1
x
2
+
3
,
{\displaystyle f(x)={\frac {1}{\sqrt {x^{2}+3}}},}
Ox , ašimi Oy ir ašiai Ox statmena tiese taške x=2. Beje, integruojant antru budu, kai x kinta nuo 0 iki 2, gauname:
∫
0
2
d
x
x
2
+
3
=
ln
|
x
+
x
2
+
3
|
|
0
2
=
ln
|
2
+
2
2
+
3
|
−
ln
|
0
+
0
2
+
3
|
=
ln
|
2
+
7
|
−
ln
|
3
|
=
{\displaystyle \int _{0}^{2}{\frac {dx}{\sqrt {x^{2}+3}}}=\ln |x+{\sqrt {x^{2}+3}}||_{0}^{2}=\ln |2+{\sqrt {2^{2}+3}}|-\ln |0+{\sqrt {0^{2}+3}}|=\ln |2+{\sqrt {7}}|-\ln |{\sqrt {3}}|=}
= 1.5359531053788889467996778565792 - 0.54930614433405484569762261846126 = 0.98664696104483410110205523811797. Į anksčiau pirmu budu gautą integralą, įstatę
t
=
x
2
+
3
x
2
=
x
2
+
3
x
,
{\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}}={\frac {\sqrt {x^{2}+3}}{x}},}
1
2
(
−
ln
(
t
−
1
)
+
ln
(
t
+
1
)
)
=
1
2
(
−
ln
(
x
2
+
3
x
−
1
)
+
ln
(
x
2
+
3
x
+
1
)
)
=
1
2
ln
x
2
+
3
x
+
1
x
2
+
3
x
−
1
=
{\displaystyle {\frac {1}{2}}(-\ln(t-1)+\ln(t+1))={\frac {1}{2}}(-\ln({\frac {\sqrt {x^{2}+3}}{x}}-1)+\ln({\frac {\sqrt {x^{2}+3}}{x}}+1))={\frac {1}{2}}\ln {\frac {{\frac {\sqrt {x^{2}+3}}{x}}+1}{{\frac {\sqrt {x^{2}+3}}{x}}-1}}=}
=
1
2
ln
x
2
+
3
+
x
x
x
2
+
3
−
x
x
=
1
2
ln
x
2
+
3
+
x
x
2
+
3
−
x
=
1
2
(
−
ln
(
x
2
+
3
−
x
)
+
ln
(
x
2
+
3
+
x
)
)
.
{\displaystyle ={\frac {1}{2}}\ln {\frac {\frac {{\sqrt {x^{2}+3}}+x}{x}}{\frac {{\sqrt {x^{2}+3}}-x}{x}}}={\frac {1}{2}}\ln {\frac {{\sqrt {x^{2}+3}}+x}{{\sqrt {x^{2}+3}}-x}}={\frac {1}{2}}(-\ln({\sqrt {x^{2}+3}}-x)+\ln({\sqrt {x^{2}+3}}+x)).}
Tokiu budu gavome, kad
∫
d
x
x
2
+
3
=
1
2
(
−
ln
(
x
2
+
3
−
x
)
+
ln
(
x
2
+
3
+
x
)
)
+
C
.
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3}}}={\frac {1}{2}}(-\ln({\sqrt {x^{2}+3}}-x)+\ln({\sqrt {x^{2}+3}}+x))+C.}
O bendru atveju gauname tokį, tikriausiai niekam nematytą, integralą:
∫
d
x
x
2
+
a
2
=
1
2
(
−
ln
(
x
2
+
a
2
−
x
)
+
ln
(
x
2
+
a
2
+
x
)
)
+
C
.
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+a^{2}}}}={\frac {1}{2}}(-\ln({\sqrt {x^{2}+a^{2}}}-x)+\ln({\sqrt {x^{2}+a^{2}}}+x))+C.}
Integruodami uždavinio sąlygos integralą nuo 0 iki 2, gauname:
∫
0
2
d
x
x
2
+
3
=
1
2
(
−
ln
(
x
2
+
3
−
x
)
+
ln
(
x
2
+
3
+
x
)
)
|
0
2
=
{\displaystyle \int _{0}^{2}{\frac {dx}{\sqrt {x^{2}+3}}}={\frac {1}{2}}(-\ln({\sqrt {x^{2}+3}}-x)+\ln({\sqrt {x^{2}+3}}+x))|_{0}^{2}=}
=
1
2
(
−
ln
(
2
2
+
3
−
2
)
+
ln
(
2
2
+
3
+
2
)
)
−
1
2
(
−
ln
(
0
2
+
3
−
0
)
+
ln
(
0
2
+
3
+
0
)
)
=
{\displaystyle ={\frac {1}{2}}(-\ln({\sqrt {2^{2}+3}}-2)+\ln({\sqrt {2^{2}+3}}+2))-{\frac {1}{2}}(-\ln({\sqrt {0^{2}+3}}-0)+\ln({\sqrt {0^{2}+3}}+0))=}
=
1
2
(
−
ln
(
7
−
2
)
+
ln
(
7
+
2
)
)
−
1
2
(
−
ln
3
+
ln
3
)
=
1
2
(
−
ln
(
7
−
2
)
+
ln
(
7
+
2
)
)
=
{\displaystyle ={\frac {1}{2}}(-\ln({\sqrt {7}}-2)+\ln({\sqrt {7}}+2))-{\frac {1}{2}}(-\ln {\sqrt {3}}+\ln {\sqrt {3}})={\frac {1}{2}}(-\ln({\sqrt {7}}-2)+\ln({\sqrt {7}}+2))=}
=
1
2
(
−
(
−
0.43734081671
)
+
1.535953105
)
=
1
2
⋅
1.973293922
=
{\displaystyle ={\frac {1}{2}}(-(-0.43734081671)+1.535953105)={\frac {1}{2}}\cdot 1.973293922=}
=0.98664696104483410110205523811797.
Toks pat atsakymas, kaip ir integruojant anksčiau. Dar galima gauti kitokia šio integralo išraišką. Štai taip:
1
2
ln
x
2
+
3
+
x
x
2
+
3
−
x
=
1
2
ln
(
x
2
+
3
+
x
)
(
x
2
+
3
+
x
)
(
x
2
+
3
−
x
)
(
x
2
+
3
+
x
)
=
1
2
ln
(
x
2
+
3
+
x
)
2
(
x
2
+
3
)
−
x
2
=
{\displaystyle {\frac {1}{2}}\ln {\frac {{\sqrt {x^{2}+3}}+x}{{\sqrt {x^{2}+3}}-x}}={\frac {1}{2}}\ln {\frac {({\sqrt {x^{2}+3}}+x)({\sqrt {x^{2}+3}}+x)}{({\sqrt {x^{2}+3}}-x)({\sqrt {x^{2}+3}}+x)}}={\frac {1}{2}}\ln {\frac {({\sqrt {x^{2}+3}}+x)^{2}}{(x^{2}+3)-x^{2}}}=}
=
1
2
ln
(
x
2
+
3
+
x
)
2
3
=
−
1
2
ln
|
3
|
+
1
2
ln
|
(
x
2
+
3
+
x
)
2
|
=
−
1
2
ln
|
3
|
+
ln
|
x
2
+
3
+
x
|
.
{\displaystyle ={\frac {1}{2}}\ln {\frac {({\sqrt {x^{2}+3}}+x)^{2}}{3}}=-{\frac {1}{2}}\ln |3|+{\frac {1}{2}}\ln |({\sqrt {x^{2}+3}}+x)^{2}|=-{\frac {1}{2}}\ln |3|+\ln |{\sqrt {x^{2}+3}}+x|.}
Tada
∫
d
x
x
2
+
a
2
=
−
ln
|
a
|
+
ln
|
x
2
+
a
2
+
x
|
+
C
.
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+a^{2}}}}=-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|+C.}
Integruodami iš pirmo būdo ką tik gautą integralą nuo 0 iki x, turėsime:
∫
0
x
d
x
x
2
+
a
2
=
(
−
ln
|
a
|
+
ln
|
x
2
+
a
2
+
x
|
)
|
0
x
=
(
−
ln
|
a
|
+
ln
|
x
2
+
a
2
+
x
|
)
−
(
−
ln
|
a
|
+
ln
|
0
2
+
a
2
+
0
|
)
=
{\displaystyle \int _{0}^{x}{\frac {dx}{\sqrt {x^{2}+a^{2}}}}=(-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|)|_{0}^{x}=(-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|)-(-\ln |a|+\ln |{\sqrt {0^{2}+a^{2}}}+0|)=}
=
(
−
ln
|
a
|
+
ln
|
x
2
+
a
2
+
x
|
)
−
(
−
ln
|
a
|
+
ln
|
a
2
|
)
=
−
ln
|
a
|
+
ln
|
x
2
+
a
2
+
x
|
.
{\displaystyle =(-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|)-(-\ln |a|+\ln |{\sqrt {a^{2}}}|)=-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|.}
Antru budu integruodami nuo 0 iki x, turime:
∫
0
x
d
x
x
2
+
a
2
=
ln
|
x
+
x
2
+
a
2
|
|
0
x
=
ln
|
x
+
x
2
+
a
2
|
−
ln
|
0
+
0
2
+
a
2
|
=
ln
|
x
+
x
2
+
a
2
|
−
ln
|
a
2
|
=
ln
|
x
+
x
2
+
a
2
|
−
ln
|
a
|
.
{\displaystyle \int _{0}^{x}{\frac {dx}{\sqrt {x^{2}+a^{2}}}}=\ln |x+{\sqrt {x^{2}+a^{2}}}||_{0}^{x}=\ln |x+{\sqrt {x^{2}+a^{2}}}|-\ln |0+{\sqrt {0^{2}+a^{2}}}|=\ln |x+{\sqrt {x^{2}+a^{2}}}|-\ln |{\sqrt {a^{2}}}|=\ln |x+{\sqrt {x^{2}+a^{2}}}|-\ln |a|.}
Gavome tokius pačius integralus ir įstačius vietoj x ir a bet kokias reikšmes, abiais būdais gausime tokias pat išraiškas ir atsakymus.
∫
1
+
x
4
3
x
d
x
=
∫
x
−
1
/
2
(
1
+
x
1
/
4
)
1
/
3
d
x
;
{\displaystyle \int {\frac {\sqrt[{3}]{1+{\sqrt[{4}]{x}}}}{\sqrt {x}}}dx=\int x^{-1/2}(1+x^{1/4})^{1/3}dx;}
m
=
−
1
2
,
n
=
1
4
,
p
=
1
3
,
m
+
1
n
=
−
1
/
2
+
1
1
/
4
=
1
/
2
1
/
4
=
2
{\displaystyle m=-{\frac {1}{2}},\;\;n={\frac {1}{4}},\;\;p={\frac {1}{3}},\;\;{\frac {m+1}{n}}={\frac {-1/2+1}{1/4}}={\frac {1/2}{1/4}}=2\;}
Keičiame
a
+
b
x
n
=
t
α
,
{\displaystyle a+bx^{n}=t^{\alpha },}
α
{\displaystyle \alpha }
p vardiklis. Tada pakeitimas yra
1
+
x
1
/
4
=
t
3
,
{\displaystyle 1+x^{1/4}=t^{3},\;}
x
1
/
4
=
t
3
−
1
,
{\displaystyle x^{1/4}=t^{3}-1,\;}
x
=
(
t
3
−
1
)
4
,
{\displaystyle x=(t^{3}-1)^{4},\;}
d
x
=
4
(
t
3
−
1
)
3
⋅
3
t
2
d
t
=
12
t
2
(
t
3
−
1
)
3
d
t
.
{\displaystyle dx=4(t^{3}-1)^{3}\cdot 3t^{2}\;dt=12t^{2}(t^{3}-1)^{3}\;dt.\;}
1
+
x
1
/
4
3
=
t
.
{\displaystyle {\sqrt[{3}]{1+x^{1/4}}}=t.}
∫
1
+
x
4
3
x
d
x
=
∫
x
−
1
/
2
(
1
+
x
1
/
4
)
1
/
3
d
x
=
∫
(
t
3
−
1
)
−
2
(
1
+
(
t
3
−
1
)
)
1
/
3
12
t
2
(
t
3
−
1
)
3
d
t
=
{\displaystyle \int {\frac {\sqrt[{3}]{1+{\sqrt[{4}]{x}}}}{\sqrt {x}}}dx=\int x^{-1/2}(1+x^{1/4})^{1/3}dx=\int (t^{3}-1)^{-2}(1+(t^{3}-1))^{1/3}12t^{2}(t^{3}-1)^{3}\;dt=}
=
∫
(
1
+
(
t
3
−
1
)
)
1
/
3
12
t
2
(
t
3
−
1
)
d
t
=
12
∫
(
t
3
)
1
/
3
t
2
(
t
3
−
1
)
d
t
=
12
∫
t
3
(
t
3
−
1
)
d
t
=
12
∫
(
t
6
−
t
3
)
d
t
=
{\displaystyle =\int (1+(t^{3}-1))^{1/3}12t^{2}(t^{3}-1)\;dt=12\int (t^{3})^{1/3}t^{2}(t^{3}-1)\;dt=12\int t^{3}(t^{3}-1)\;dt=12\int (t^{6}-t^{3})\;dt=}
=
12
(
t
7
7
−
t
4
4
)
+
C
=
3
7
t
4
(
4
t
3
−
7
)
+
C
=
3
7
(
1
+
x
1
/
4
3
)
4
(
4
(
1
+
x
1
/
4
)
−
7
)
+
C
=
3
7
1
+
x
1
/
4
3
(
1
+
x
1
/
4
)
(
4
(
1
+
x
1
/
4
)
−
7
)
+
C
.
{\displaystyle =12({\frac {t^{7}}{7}}-{\frac {t^{4}}{4}})+C={\frac {3}{7}}t^{4}(4t^{3}-7)+C={\frac {3}{7}}({\sqrt[{3}]{1+x^{1/4}}})^{4}(4(1+x^{1/4})-7)+C={\frac {3}{7}}{\sqrt[{3}]{1+x^{1/4}}}(1+x^{1/4})(4(1+x^{1/4})-7)+C.}
I
=
∫
d
x
x
2
a
+
b
x
2
=
∫
x
−
2
(
a
+
b
x
2
)
−
1
2
d
x
.
{\displaystyle I=\int {\frac {dx}{x^{2}{\sqrt {a+bx^{2}}}}}=\int x^{-2}(a+bx^{2})^{-{\frac {1}{2}}}dx.}
Šiame pavyzdyje
m
=
−
2
,
n
=
2
,
p
=
−
1
2
,
{\displaystyle m=-2,\;n=2,\;p=-{\frac {1}{2}},}
m
+
1
n
+
p
=
−
2
+
1
2
−
1
2
=
−
1
2
−
1
2
=
−
1
{\displaystyle {\frac {m+1}{n}}+p={\frac {-2+1}{2}}-{\frac {1}{2}}={\frac {-1}{2}}-{\frac {1}{2}}=-1\;}
Tada
a
+
b
x
n
=
t
α
x
n
,
{\displaystyle a+bx^{n}=t^{\alpha }x^{n},}
α
{\displaystyle \alpha }
p vardiklis;
a
+
b
x
2
=
t
2
x
2
.
{\displaystyle a+bx^{2}=t^{2}x^{2}.}
t
=
a
x
2
+
b
;
t
2
x
2
−
b
x
2
=
a
,
(
t
2
−
b
)
x
2
=
a
,
x
=
a
t
2
−
b
;
d
x
=
−
1
2
a
2
t
d
t
(
t
2
−
b
)
3
/
2
=
−
a
t
d
t
(
t
2
−
b
)
3
.
{\displaystyle t={\sqrt {{\frac {a}{x^{2}}}+b}};\;\;t^{2}x^{2}-bx^{2}=a,\;(t^{2}-b)x^{2}=a,\;x={\frac {\sqrt {a}}{\sqrt {t^{2}-b}}};\;\;dx=-{\frac {1}{2}}{\frac {{\sqrt {a}}\;2t\;dt}{(t^{2}-b)^{3/2}}}={\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}.}
Tada gauname
I
=
∫
x
−
2
(
a
+
b
x
2
)
−
1
2
d
x
=
∫
t
2
−
b
a
(
a
+
b
a
t
2
−
b
)
−
1
2
−
a
t
d
t
(
t
2
−
b
)
3
=
∫
t
2
−
b
a
(
a
(
t
2
−
b
)
+
a
b
t
2
−
b
)
−
1
2
−
a
t
d
t
(
t
2
−
b
)
3
=
{\displaystyle I=\int x^{-2}(a+bx^{2})^{-{\frac {1}{2}}}dx=\int {\frac {t^{2}-b}{a}}(a+b{\frac {a}{t^{2}-b}})^{-{\frac {1}{2}}}{\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}=\int {\frac {t^{2}-b}{a}}({\frac {a(t^{2}-b)+ab}{t^{2}-b}})^{-{\frac {1}{2}}}{\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}=}
=
∫
t
2
−
b
a
(
t
2
t
2
−
b
)
−
1
2
−
a
t
d
t
(
t
2
−
b
)
3
=
∫
t
2
−
b
a
t
2
−
b
t
−
a
t
d
t
(
t
2
−
b
)
3
=
−
∫
d
t
a
=
{\displaystyle =\int {\frac {t^{2}-b}{a}}({\frac {t^{2}}{t^{2}-b}})^{-{\frac {1}{2}}}{\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}=\int {\frac {t^{2}-b}{a}}{\frac {\sqrt {t^{2}-b}}{t}}{\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}=-\int {\frac {dt}{\sqrt {a}}}=}
=
−
t
a
+
C
=
−
a
x
2
+
b
a
+
C
.
{\displaystyle =-{\frac {t}{\sqrt {a}}}+C=-{\frac {\sqrt {{\frac {a}{x^{2}}}+b}}{\sqrt {a}}}+C.}
Apskaičiuosime integralą
I
=
∫
x
5
(
1
−
x
2
)
−
1
2
d
x
.
{\displaystyle I=\int x^{5}(1-x^{2})^{-{\frac {1}{2}}}dx.}
m
=
5
,
n
=
2
,
p
=
−
1
2
,
{\displaystyle m=5,\;n=2,\;p=-{\frac {1}{2}},}
m
+
1
n
=
3
{\displaystyle {\frac {m+1}{n}}=3\;}
t
=
1
−
x
2
,
t
2
=
1
−
x
2
,
x
=
1
−
t
2
,
d
x
=
−
2
t
d
t
2
1
−
t
2
=
−
t
d
t
1
−
t
2
,
{\displaystyle t={\sqrt {1-x^{2}}},\;t^{2}=1-x^{2},\;x={\sqrt {1-t^{2}}},\;\;dx={\frac {-2t\;dt}{2{\sqrt {1-t^{2}}}}}={\frac {-t\;dt}{\sqrt {1-t^{2}}}},}
gauname
I
=
∫
x
5
(
1
−
x
2
)
−
1
2
d
x
=
∫
(
1
−
t
2
)
5
(
1
−
[
1
−
t
2
]
)
−
1
2
−
t
d
t
1
−
t
2
=
∫
1
−
t
2
(
1
−
t
2
)
2
(
t
2
)
−
1
2
−
t
d
t
1
−
t
2
=
{\displaystyle I=\int x^{5}(1-x^{2})^{-{\frac {1}{2}}}dx=\int ({\sqrt {1-t^{2}}})^{5}(1-[1-t^{2}])^{-{\frac {1}{2}}}{\frac {-t\;dt}{\sqrt {1-t^{2}}}}=\int {\sqrt {1-t^{2}}}(1-t^{2})^{2}(t^{2})^{-{\frac {1}{2}}}{\frac {-t\;dt}{\sqrt {1-t^{2}}}}=}
=
∫
1
−
t
2
(
1
−
t
2
)
2
1
t
−
t
d
t
1
−
t
2
=
−
∫
(
1
−
t
2
)
2
d
t
=
−
∫
(
1
−
2
t
2
+
t
4
)
d
t
=
−
∫
d
t
+
2
∫
t
2
d
t
−
t
4
d
t
=
{\displaystyle =\int {\sqrt {1-t^{2}}}(1-t^{2})^{2}{\frac {1}{t}}{\frac {-t\;dt}{\sqrt {1-t^{2}}}}=-\int (1-t^{2})^{2}\;dt=-\int (1-2t^{2}+t^{4})dt=-\int dt+2\int t^{2}\;dt-t^{4}\;dt=}
=
−
t
+
2
3
t
3
−
t
5
5
+
C
=
−
1
−
x
2
+
2
3
(
1
−
x
2
)
3
−
(
1
−
x
2
)
5
5
+
C
.
{\displaystyle =-t+{\frac {2}{3}}t^{3}-{\frac {t^{5}}{5}}+C=-{\sqrt {1-x^{2}}}+{\frac {2}{3}}{\sqrt {(1-x^{2})^{3}}}-{\frac {\sqrt {(1-x^{2})^{5}}}{5}}+C.}
![{\displaystyle \int {\frac {\sqrt {x}}{x^{\frac {3}{4}}+1}}dx=\int {\frac {u^{2}\cdot 4u^{3}du}{u^{3}+1}}=4[\int u^{2}du-\int {\frac {u^{2}du}{u^{3}+1}}]=4[{\frac {u^{3}}{3}}-{\frac {1}{3}}\int {\frac {d(u^{3}+1)}{u^{3}+1}}]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fefd470530942312874211bf05d5f9ad9a1f67c)
![{\displaystyle ={\frac {4}{3}}[u^{3}-\ln(u^{3}+1)]+C={\frac {4}{3}}[x^{\frac {3}{4}}-\ln(x^{\frac {3}{4}}+1)]+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cbc23a01bbd1a78f7aed878d27f3fd7402bda69)
![{\displaystyle \int {\frac {dx}{x+{\sqrt {x^{2}+x+1}}}}=\int [{\frac {2}{t}}-{\frac {3}{1+2t}}-{\frac {3}{(1+2t)^{2}}}]dt=2\int {\frac {dt}{t}}-{\frac {3}{2}}\int {\frac {d(1+2t)}{1+2t}}-{\frac {3}{2}}\int {\frac {d(1+2t)}{(1+2t)^{2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/628a255a96d7f5937508df3007da061cdf125968)
![{\displaystyle \int {\frac {\sqrt[{3}]{1+{\sqrt[{4}]{x}}}}{\sqrt {x}}}dx=\int x^{-1/2}(1+x^{1/4})^{1/3}dx;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08393264fd701a0223068184ae0168c0701a8774)
![{\displaystyle {\sqrt[{3}]{1+x^{1/4}}}=t.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f765f9ed01175aaabae58702d58804583c311d0e)
![{\displaystyle \int {\frac {\sqrt[{3}]{1+{\sqrt[{4}]{x}}}}{\sqrt {x}}}dx=\int x^{-1/2}(1+x^{1/4})^{1/3}dx=\int (t^{3}-1)^{-2}(1+(t^{3}-1))^{1/3}12t^{2}(t^{3}-1)^{3}\;dt=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d2382f7f6b604ced2ead0aaf628d4364e0859cb)
![{\displaystyle =12({\frac {t^{7}}{7}}-{\frac {t^{4}}{4}})+C={\frac {3}{7}}t^{4}(4t^{3}-7)+C={\frac {3}{7}}({\sqrt[{3}]{1+x^{1/4}}})^{4}(4(1+x^{1/4})-7)+C={\frac {3}{7}}{\sqrt[{3}]{1+x^{1/4}}}(1+x^{1/4})(4(1+x^{1/4})-7)+C.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b521a7108e9e5377166e3607a6e185eb009d79d0)
![{\displaystyle I=\int x^{5}(1-x^{2})^{-{\frac {1}{2}}}dx=\int ({\sqrt {1-t^{2}}})^{5}(1-[1-t^{2}])^{-{\frac {1}{2}}}{\frac {-t\;dt}{\sqrt {1-t^{2}}}}=\int {\sqrt {1-t^{2}}}(1-t^{2})^{2}(t^{2})^{-{\frac {1}{2}}}{\frac {-t\;dt}{\sqrt {1-t^{2}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34b4fbaf99f3eb0a6334733583ad9013c69b053a)
