∫
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.}
1
x
2
+
3
=
1
4
2
+
3
=
1
19
=
1
4.358898944
=
0.229415733.
{\displaystyle {\frac {1}{\sqrt {x^{2}+3}}}={\frac {1}{\sqrt {4^{2}+3}}}={\frac {1}{\sqrt {19}}}={\frac {1}{4.358898944}}=0.229415733.}
(
ln
|
x
+
x
2
+
3
|
+
C
)
′
=
(
x
+
x
2
+
3
)
′
x
+
x
2
+
3
+
C
′
=
1
+
1
2
⋅
(
x
2
+
3
)
′
x
2
+
3
x
+
x
2
+
3
=
1
+
2
x
2
x
2
+
3
x
+
x
2
+
3
=
1
+
x
x
2
+
3
x
+
x
2
+
3
=
{\displaystyle (\ln |x+{\sqrt {x^{2}+3}}|+C)'={\frac {(x+{\sqrt {x^{2}+3}})'}{x+{\sqrt {x^{2}+3}}}}+C'={\frac {1+{1 \over 2}\cdot {\frac {(x^{2}+3)'}{\sqrt {x^{2}+3}}}}{x+{\sqrt {x^{2}+3}}}}={\frac {1+{\frac {2x}{2{\sqrt {x^{2}+3}}}}}{x+{\sqrt {x^{2}+3}}}}={\frac {1+{\frac {x}{\sqrt {x^{2}+3}}}}{x+{\sqrt {x^{2}+3}}}}=}
=
1
+
4
4
2
+
3
4
+
4
2
+
3
=
1
+
4
19
4
+
19
=
1
+
4
4.358898944
4
+
4.358898944
=
1
+
0.917662935
8.358898944
=
1.917662935
8.358898944
=
0.229415733.
{\displaystyle ={\frac {1+{\frac {4}{\sqrt {4^{2}+3}}}}{4+{\sqrt {4^{2}+3}}}}={\frac {1+{\frac {4}{\sqrt {19}}}}{4+{\sqrt {19}}}}={\frac {1+{\frac {4}{4.358898944}}}{4+4.358898944}}={\frac {1+0.917662935}{8.358898944}}={\frac {1.917662935}{8.358898944}}=0.229415733.}
∫
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.}
Patikriname ar atsakymai bus vienodi istacius
x
=
3
{\displaystyle x=3}
(
30
x
−
2
27
5
−
x
−
12
27
5
−
x
x
−
2
+
C
)
′
=
30
27
⋅
1
2
⋅
5
−
x
x
−
2
⋅
(
x
−
2
5
−
x
)
′
−
12
27
⋅
1
2
⋅
x
−
2
5
−
x
⋅
(
5
−
x
x
−
2
)
′
=
{\displaystyle ({\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {5-x}{x-2}}}+C)'={\frac {30}{27}}\cdot {\frac {1}{2}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot ({\frac {x-2}{5-x}})'-{\frac {12}{27}}\cdot {\frac {1}{2}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot ({\frac {5-x}{x-2}})'=}
=
15
27
⋅
5
−
x
x
−
2
⋅
x
(
5
−
x
)
−
(
x
−
2
)
⋅
(
−
x
)
(
5
−
x
)
2
−
6
27
⋅
x
−
2
5
−
x
⋅
−
x
(
x
−
2
)
−
(
5
−
x
)
x
(
x
−
2
)
2
=
{\displaystyle ={\frac {15}{27}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot {\frac {x(5-x)-(x-2)\cdot (-x)}{(5-x)^{2}}}-{\frac {6}{27}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot {\frac {-x(x-2)-(5-x)x}{(x-2)^{2}}}=}
=
15
27
⋅
5
−
x
x
−
2
⋅
5
x
−
x
2
+
x
2
−
2
x
(
5
−
x
)
2
−
6
27
⋅
x
−
2
5
−
x
⋅
−
x
2
+
2
x
−
5
x
+
x
2
(
x
−
2
)
2
=
{\displaystyle ={\frac {15}{27}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot {\frac {5x-x^{2}+x^{2}-2x}{(5-x)^{2}}}-{\frac {6}{27}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot {\frac {-x^{2}+2x-5x+x^{2}}{(x-2)^{2}}}=}
=
15
27
⋅
5
−
x
x
−
2
⋅
3
x
(
5
−
x
)
2
−
6
27
⋅
x
−
2
5
−
x
⋅
−
3
x
(
x
−
2
)
2
=
1
27
(
5
−
3
3
−
2
⋅
15
⋅
3
⋅
3
(
5
−
3
)
2
+
3
−
2
5
−
3
⋅
18
⋅
3
(
3
−
2
)
2
)
=
{\displaystyle ={\frac {15}{27}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot {\frac {3x}{(5-x)^{2}}}-{\frac {6}{27}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot {\frac {-3x}{(x-2)^{2}}}={\frac {1}{27}}({\sqrt {\frac {5-3}{3-2}}}\cdot {\frac {15\cdot 3\cdot 3}{(5-3)^{2}}}+{\frac {\sqrt {3-2}}{\sqrt {5-3}}}\cdot {\frac {18\cdot 3}{(3-2)^{2}}})=}
=
1
27
(
2
1
⋅
135
2
2
+
1
2
⋅
54
1
2
)
=
1
27
(
135
2
4
+
54
2
)
=
1
27
(
47.72970773
+
38.18376618
)
=
3.181980515.
{\displaystyle ={\frac {1}{27}}({\sqrt {\frac {2}{1}}}\cdot {\frac {135}{2^{2}}}+{\frac {\sqrt {1}}{\sqrt {2}}}\cdot {\frac {54}{1^{2}}})={\frac {1}{27}}({\frac {135{\sqrt {2}}}{4}}+{\frac {54}{\sqrt {2}}})={\frac {1}{27}}(47.72970773+38.18376618)=3.181980515.}
x
(
7
x
−
10
−
x
2
)
3
=
3
(
7
⋅
3
−
10
−
3
2
)
3
=
3
(
21
−
10
−
9
)
3
=
3
2
3
=
3
8
=
1.060660172.
{\displaystyle {\frac {x}{\sqrt {(7x-10-x^{2})^{3}}}}={\frac {3}{\sqrt {(7\cdot 3-10-3^{2})^{3}}}}={\frac {3}{\sqrt {(21-10-9)^{3}}}}={\frac {3}{\sqrt {2^{3}}}}={\frac {3}{\sqrt {8}}}=1.060660172.}
Pabandome gauti teisinga atsakyma isvestine darant kitokiu budu (ir istacius x=3), nors sis budas ligtais 95%, kad neteisingas:
(
30
x
−
2
27
5
−
x
−
12
27
5
−
x
x
−
2
+
C
)
′
=
30
27
⋅
5
−
x
2
x
−
2
−
x
−
2
2
5
−
x
(
5
−
x
)
2
−
12
27
⋅
x
−
2
2
5
−
x
−
5
−
x
2
x
−
2
(
x
−
2
)
2
=
{\displaystyle ({\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {5-x}{x-2}}}+C)'={\frac {30}{27}}\cdot {\frac {{\frac {\sqrt {5-x}}{2{\sqrt {x-2}}}}-{\frac {\sqrt {x-2}}{2{\sqrt {5-x}}}}}{({\sqrt {5-x}})^{2}}}-{\frac {12}{27}}\cdot {\frac {{\frac {\sqrt {x-2}}{2{\sqrt {5-x}}}}-{\frac {\sqrt {5-x}}{2{\sqrt {x-2}}}}}{({\sqrt {x-2}})^{2}}}=}
=
30
27
⋅
5
−
3
2
3
−
2
−
3
−
2
2
5
−
3
(
5
−
3
)
2
−
12
27
⋅
3
−
2
2
5
−
3
−
5
−
3
2
3
−
2
(
3
−
2
)
2
=
30
27
⋅
2
2
1
−
1
2
2
(
2
)
2
−
12
27
⋅
1
2
2
−
2
2
1
(
1
)
2
=
{\displaystyle ={\frac {30}{27}}\cdot {\frac {{\frac {\sqrt {5-3}}{2{\sqrt {3-2}}}}-{\frac {\sqrt {3-2}}{2{\sqrt {5-3}}}}}{({\sqrt {5-3}})^{2}}}-{\frac {12}{27}}\cdot {\frac {{\frac {\sqrt {3-2}}{2{\sqrt {5-3}}}}-{\frac {\sqrt {5-3}}{2{\sqrt {3-2}}}}}{({\sqrt {3-2}})^{2}}}={\frac {30}{27}}\cdot {\frac {{\frac {\sqrt {2}}{2{\sqrt {1}}}}-{\frac {\sqrt {1}}{2{\sqrt {2}}}}}{({\sqrt {2}})^{2}}}-{\frac {12}{27}}\cdot {\frac {{\frac {\sqrt {1}}{2{\sqrt {2}}}}-{\frac {\sqrt {2}}{2{\sqrt {1}}}}}{({\sqrt {1}})^{2}}}=}
=
30
27
⋅
2
−
1
2
2
2
−
12
27
⋅
1
−
2
2
2
=
30
27
⋅
1
4
2
+
12
27
⋅
1
2
2
=
15
54
2
+
6
27
2
=
15
−
12
54
2
=
3
54
2
=
1
18
2
=
0.03928371.
{\displaystyle ={\frac {30}{27}}\cdot {\frac {\frac {2-1}{2{\sqrt {2}}}}{2}}-{\frac {12}{27}}\cdot {\frac {1-2}{2{\sqrt {2}}}}={\frac {30}{27}}\cdot {\frac {1}{4{\sqrt {2}}}}+{\frac {12}{27}}\cdot {\frac {1}{2{\sqrt {2}}}}={\frac {15}{54{\sqrt {2}}}}+{\frac {6}{27{\sqrt {2}}}}={\frac {15-12}{54{\sqrt {2}}}}={\frac {3}{54{\sqrt {2}}}}={\frac {1}{18{\sqrt {2}}}}=0.03928371.}
(
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)'=}
=
30
27
(
1
2
⋅
(
x
−
2
)
′
x
−
2
5
−
x
+
−
x
−
2
⋅
(
5
−
x
)
′
(
5
−
x
)
2
)
−
12
27
(
1
2
⋅
(
5
−
x
)
′
5
−
x
x
−
2
+
−
5
−
x
⋅
(
x
−
2
)
′
(
x
−
2
)
2
)
=
{\displaystyle ={\frac {30}{27}}({\frac {1}{2}}\cdot {\frac {(x-2)'}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {-{\sqrt {x-2}}\cdot ({\sqrt {5-x}})'}{({\sqrt {5-x}})^{2}}})-{\frac {12}{27}}({\frac {1}{2}}\cdot {\frac {(5-x)'}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {-{\sqrt {5-x}}\cdot ({\sqrt {x-2}})'}{({\sqrt {x-2}})^{2}}})=}
=
30
27
(
1
2
⋅
1
x
−
2
5
−
x
+
−
x
−
2
⋅
1
2
⋅
(
5
−
x
)
′
5
−
x
⋅
(
5
−
x
)
)
−
12
27
(
1
2
⋅
−
1
5
−
x
x
−
2
+
−
5
−
x
⋅
1
2
(
x
−
2
)
′
x
−
2
(
x
−
2
)
)
=
{\displaystyle ={\frac {30}{27}}({\frac {1}{2}}\cdot {\frac {1}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {-{\sqrt {x-2}}\cdot {\frac {1}{2}}\cdot (5-x)'}{{\sqrt {5-x}}\cdot (5-x)}})-{\frac {12}{27}}({\frac {1}{2}}\cdot {\frac {-1}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {-{\sqrt {5-x}}\cdot {\frac {1}{2}}(x-2)'}{{\sqrt {x-2}}(x-2)}})=}
=
15
27
(
1
x
−
2
5
−
x
+
−
x
−
2
⋅
(
−
1
)
5
−
x
⋅
(
5
−
x
)
)
−
6
27
(
−
1
5
−
x
x
−
2
+
−
5
−
x
⋅
1
x
−
2
(
x
−
2
)
)
=
{\displaystyle ={\frac {15}{27}}({\frac {1}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {-{\sqrt {x-2}}\cdot (-1)}{{\sqrt {5-x}}\cdot (5-x)}})-{\frac {6}{27}}({\frac {-1}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {-{\sqrt {5-x}}\cdot 1}{{\sqrt {x-2}}(x-2)}})=}
=
15
27
(
1
x
−
2
5
−
x
+
x
−
2
5
−
x
⋅
(
5
−
x
)
)
+
6
27
(
1
5
−
x
x
−
2
+
5
−
x
x
−
2
(
x
−
2
)
)
=
{\displaystyle ={\frac {15}{27}}({\frac {1}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {\sqrt {x-2}}{{\sqrt {5-x}}\cdot (5-x)}})+{\frac {6}{27}}({\frac {1}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {\sqrt {5-x}}{{\sqrt {x-2}}(x-2)}})=}
=
15
27
(
1
3
−
2
5
−
3
+
3
−
2
5
−
3
⋅
(
5
−
3
)
)
+
6
27
(
1
5
−
3
3
−
2
+
5
−
3
3
−
2
(
3
−
2
)
)
=
{\displaystyle ={\frac {15}{27}}({\frac {1}{{\sqrt {3-2}}{\sqrt {5-3}}}}+{\frac {\sqrt {3-2}}{{\sqrt {5-3}}\cdot (5-3)}})+{\frac {6}{27}}({\frac {1}{{\sqrt {5-3}}{\sqrt {3-2}}}}+{\frac {\sqrt {5-3}}{{\sqrt {3-2}}(3-2)}})=}
=
15
27
(
1
2
+
1
2
⋅
2
)
+
6
27
(
1
2
+
2
1
)
=
15
27
⋅
2
+
1
2
2
+
6
27
⋅
1
+
2
2
=
45
54
2
+
18
27
2
=
5
6
2
+
2
3
2
=
{\displaystyle ={\frac {15}{27}}({\frac {1}{\sqrt {2}}}+{\frac {1}{{\sqrt {2}}\cdot 2}})+{\frac {6}{27}}({\frac {1}{\sqrt {2}}}+{\frac {\sqrt {2}}{1}})={\frac {15}{27}}\cdot {\frac {2+1}{2{\sqrt {2}}}}+{\frac {6}{27}}\cdot {\frac {1+2}{\sqrt {2}}}={\frac {45}{54{\sqrt {2}}}}+{\frac {18}{27{\sqrt {2}}}}={\frac {5}{6{\sqrt {2}}}}+{\frac {2}{3{\sqrt {2}}}}=}
=
0.589255651
+
0.47140452
=
1.060660172.
{\displaystyle =0.589255651+0.47140452=1.060660172.}
Sis budas pasirode esas teisingas. Cia buvo pasinaudota dviejomis formulemis:
(
f
g
)
′
=
f
′
g
+
f
g
′
;
{\displaystyle \left({fg}\right)'=f'g+fg';}
(
1
f
)
′
=
−
f
′
f
2
,
f
≠
0.
{\displaystyle \left({\frac {1}{f}}\right)'={\frac {-f'}{f^{2}}},\qquad f\neq 0.}
Diferencijavimas vyko tokiu budu, kad pavyzdiui, sioje funkcijoje
5
−
x
x
−
2
{\displaystyle {\sqrt {\frac {5-x}{x-2}}}}
f
(
x
)
=
f
=
5
−
x
{\displaystyle f(x)=f={\sqrt {5-x}}}
g
(
x
)
=
g
=
1
x
−
2
{\displaystyle g(x)=g={\frac {1}{\sqrt {x-2}}}}
(
f
(
x
)
)
′
=
f
′
=
1
2
5
−
x
{\displaystyle (f(x))'=f'={\frac {1}{2{\sqrt {5-x}}}}}
(
g
(
x
)
)
′
=
g
′
=
−
(
x
−
2
)
′
(
x
−
2
)
2
=
−
1
2
x
−
2
(
x
−
2
)
2
.
{\displaystyle (g(x))'=g'={\frac {-({\sqrt {x-2}})'}{({\sqrt {x-2}})^{2}}}={\frac {-1}{2{\sqrt {x-2}}({\sqrt {x-2}})^{2}}}.}
Diferencialinių binomų integravimas
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
∫
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
d
t
=
−
3
2
(
t
2
−
1
)
3
d
t
.
{\displaystyle dx=-{\frac {\sqrt {3}}{2}}\cdot (t^{2}-1)^{-{\frac {3}{2}}}dt=-{\frac {\sqrt {3}}{2{\sqrt {(t^{2}-1)^{3}}}}}dt.}
∫
d
x
x
2
+
3
=
−
3
2
∫
d
t
(
t
2
−
1
)
3
⋅
3
t
2
−
1
+
3
=
−
3
2
3
∫
d
t
(
t
2
−
1
)
2
+
(
t
2
−
1
)
3
=
{\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3}}}=-{\frac {\sqrt {3}}{2}}\int {\frac {dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {{\frac {3}{t^{2}-1}}+3}}}}=-{\frac {\sqrt {3}}{2{\sqrt {3}}}}\int {\frac {dt}{\sqrt {(t^{2}-1)^{2}+(t^{2}-1)^{3}}}}=}
=
−
1
2
∫
d
t
(
t
4
−
2
t
2
+
1
)
+
(
t
6
−
3
(
t
2
)
2
+
3
t
2
−
1
3
)
=
−
1
2
∫
d
t
t
6
−
2
t
4
+
t
2
=
{\displaystyle =-{\frac {1}{2}}\int {\frac {dt}{\sqrt {(t^{4}-2t^{2}+1)+(t^{6}-3(t^{2})^{2}+3t^{2}-1^{3})}}}=-{\frac {1}{2}}\int {\frac {dt}{\sqrt {t^{6}-2t^{4}+t^{2}}}}=}
=
−
1
2
∫
d
t
t
t
4
−
2
t
2
+
1
=
−
1
2
∫
d
t
t
(
t
2
−
1
)
2
=
−
1
2
∫
d
t
t
(
t
2
−
1
)
=
−
1
2
∫
(
t
t
2
−
1
−
1
t
)
d
t
=
{\displaystyle =-{\frac {1}{2}}\int {\frac {dt}{t{\sqrt {t^{4}-2t^{2}+1}}}}=-{\frac {1}{2}}\int {\frac {dt}{t{\sqrt {(t^{2}-1)^{2}}}}}=-{\frac {1}{2}}\int {\frac {dt}{t(t^{2}-1)}}=-{\frac {1}{2}}\int {\Big (}{\frac {t}{t^{2}-1}}-{\frac {1}{t}}{\Big )}dt=}
=
−
1
4
(
ln
(
t
2
−
1
)
−
2
ln
t
)
.
{\displaystyle =-{1 \over 4}(\ln(t^{2}-1)-2\ln t).}
t
=
x
2
+
3
x
2
.
{\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}}.}
Iš tikro
d
x
=
−
3
2
⋅
(
t
2
−
1
)
−
3
2
2
t
d
t
=
−
3
(
t
2
−
1
)
3
t
d
t
.
{\displaystyle dx=-{\frac {\sqrt {3}}{2}}\cdot (t^{2}-1)^{-{\frac {3}{2}}}2t\;dt=-{\frac {\sqrt {3}}{\sqrt {(t^{2}-1)^{3}}}}t\;dt.}
∫
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
3
∫
t
d
t
(
t
2
−
1
)
2
+
(
t
2
−
1
)
3
=
{\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}}}}=-{\frac {\sqrt {3}}{\sqrt {3}}}\int {\frac {t\;dt}{\sqrt {(t^{2}-1)^{2}+(t^{2}-1)^{3}}}}=}
=
−
∫
t
d
t
(
t
4
−
2
t
2
+
1
)
+
(
t
6
−
3
(
t
2
)
2
+
3
t
2
−
1
3
)
=
−
∫
t
d
t
t
6
−
2
t
4
+
t
2
=
{\displaystyle =-\int {\frac {t\;dt}{\sqrt {(t^{4}-2t^{2}+1)+(t^{6}-3(t^{2})^{2}+3t^{2}-1^{3})}}}=-\int {\frac {t\;dt}{\sqrt {t^{6}-2t^{4}+t^{2}}}}=}
=
−
∫
t
d
t
t
t
4
−
2
t
2
+
1
=
−
∫
d
t
(
t
2
−
1
)
2
=
−
∫
d
t
t
2
−
1
=
−
∫
d
t
(
t
−
1
)
(
t
+
1
)
;
{\displaystyle =-\int {\frac {t\;dt}{t{\sqrt {t^{4}-2t^{2}+1}}}}=-\int {\frac {dt}{\sqrt {(t^{2}-1)^{2}}}}=-\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}};}
Kur
t
=
x
2
+
3
x
2
.
{\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}}.}
Bet tokia funkcija integruojama lengviau kitaip (ne per diferencialinius binomus) 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.}
Klaida čia:
∫
d
x
x
2
+
3
=
−
3
∫
t
d
t
(
t
2
−
1
)
3
⋅
3
t
2
−
1
+
3
=
−
3
3
∫
t
d
t
(
t
2
−
1
)
2
+
(
t
2
−
1
)
3
=
{\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}}}}=-{\frac {\sqrt {3}}{\sqrt {3}}}\int {\frac {t\;dt}{\sqrt {(t^{2}-1)^{2}+(t^{2}-1)^{3}}}}=}
Paraboloid (aptarimas ) 19:55, 31 gruodžio 2022 (UTC) Atsakyti
Pasirodo, kad teisingai skaičiuojant gaunamas toks pat integralas
−
∫
d
t
t
2
−
1
.
{\displaystyle -\int {\frac {dt}{t^{2}-1}}.}
−
∫
d
t
t
2
−
1
{\displaystyle -\int {\frac {dt}{t^{2}-1}}}
Dabar atrodo supratau, kaip ten skaičiuota (ir ten teisingai skaičiuota)... 
![{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\mbox{(for }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C&{\mbox{(for }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{2ax+b}}+C&{\mbox{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67fe77e22d3b8fa1d879c1dc632ff790210041f7)
![{\displaystyle \int {\sqrt {1-6x-9x^{2}}}\;dx={\frac {1}{25}}\int (u^{2}+3u-5)du={\frac {1}{25}}[{u^{3} \over 3}+{3u^{2} \over 2}-5u+C={(-5x-3)^{3} \over 3}+{3(-5x-3)^{2} \over 2}-5(-5x-3)]+C=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97ae6a588017f3088e157cd3e601d8c9c1ffeeb7)
![{\displaystyle ={\frac {1}{25}}[{(-5x)^{3}-3\cdot (-5x)^{2}\cdot 3+3\cdot (-5x)\cdot 3^{2}-3^{3} \over 3}+{3[(-5x)^{2}-2\cdot (-5x)\cdot 3+3^{2}] \over 2}+25x+15]+C=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06e81fe65333c067a0062506a190d0bf4269efdd)
![{\displaystyle ={\frac {1}{25}}[{-125x^{3}-9\cdot 25x^{2}+27\cdot (-5x)-27 \over 3}+{3[25x^{2}-6\cdot (-5x)+9] \over 2}+25x+15]+C=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77d383b3e7874b0c8a3fc2c89a88e3fb2410aae8)
![{\displaystyle ={\frac {1}{25}}[{-125x^{3}-225x^{2}-135x-27 \over 3}+{3[25x^{2}+30x+9] \over 2}+25x+15]+C=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f16be3006cea2520280845aaaf11ce63dc6b8c90)
![{\displaystyle ={\frac {1}{25}}[{-125x^{3} \over 3}-75x^{2}-45x-9+37.5x^{2}+45x+13.5+25x+15]+C={\frac {1}{25}}[{-125x^{3} \over 3}-37.5x^{2}+25x+19.5]+C.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7e39997eb1a33b3d6d4663fc6b234230ea3b712)
![{\displaystyle {\frac {1}{25}}({-125x^{3} \over 3}-37.5x^{2}+25x+19.5+C)'={\frac {1}{25}}[-125x^{2}-75x+25]={\frac {1}{25}}[-125\cdot 0.1^{2}-75\cdot 0.1+25]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/376c4a95f928cc7524cc88a9b1f49724d65c9d70)
![{\displaystyle ={\frac {1}{25}}[-1.25-7.5+25]={\frac {1}{25}}\cdot 16.25=0.65.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1eebf2ad716365ee02d7e78d42d1ee067e0f234e)
![{\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)
