Iracionaliųjų funkcijų integravimas (papildomai)
2. Trupmeninių tiesinių iracionalumų integravimas
keisti
Šiame skirsnyje įrodysime, kad bet kokios funkcijos
R
(
x
,
a
x
+
b
c
x
+
d
n
)
(
7.64
)
{\displaystyle R{\Big (}x,\;{\sqrt[{n}]{\frac {ax+b}{cx+d}}}{\Big )}\quad (7.64)}
(a , b , c ir d - realūs skačiai, n - natūrinis skaičius) integralas yra elementarioji funkcija. Tokio tipo funkcijas vadinsime trupmeniniais tiesiniais iracionalumais .
Įsitikinsime, kad (7.64) funkcijos integralas, kai
a
d
−
b
c
≠
0
{\displaystyle ad-bc\neq 0\;}
a
d
−
b
c
=
0
{\displaystyle ad-bc=0}
k ), racionalizuojamas keitiniu
t
=
a
x
+
b
c
x
+
d
n
.
{\displaystyle t={\sqrt[{n}]{\frac {ax+b}{cx+d}}}.}
t
n
=
a
x
+
b
c
x
+
d
,
t
n
(
c
x
+
d
)
=
a
x
+
b
,
t
n
c
x
−
a
x
=
b
−
d
t
n
,
(
t
n
c
−
a
)
x
=
b
−
d
t
n
,
x
=
b
−
d
t
n
t
n
c
−
a
=
d
t
n
−
b
a
−
c
t
n
,
{\displaystyle t^{n}={\frac {ax+b}{cx+d}},\;\;t^{n}(cx+d)=ax+b,\;\;t^{n}cx-ax=b-dt^{n},\;\;(t^{n}c-a)x=b-dt^{n},\;\;x={\frac {b-dt^{n}}{t^{n}c-a}}={\frac {dt^{n}-b}{a-ct^{n}}},\;\;}
d
x
=
d
n
t
n
−
1
(
a
−
c
t
n
)
−
(
d
t
n
−
b
)
(
−
c
n
t
n
−
1
)
(
a
−
c
t
n
)
2
d
t
=
(
a
d
−
b
c
)
n
t
n
−
1
(
a
−
c
t
n
)
2
d
t
.
{\displaystyle dx={\frac {dnt^{n-1}(a-ct^{n})-(dt^{n}-b)(-cnt^{n-1})}{(a-ct^{n})^{2}}}\;dt={\frac {(ad-bc)nt^{n-1}}{(a-ct^{n})^{2}}}\;dt.}
Todėl
∫
R
(
x
,
a
x
+
b
c
x
+
d
n
)
d
x
=
∫
R
(
d
t
n
−
b
a
−
c
t
n
,
t
)
(
a
d
−
b
c
)
n
t
n
−
1
(
a
−
c
t
n
)
2
d
t
.
{\displaystyle \int R{\Big (}x,\;{\sqrt[{n}]{\frac {ax+b}{cx+d}}}{\Big )}\;dx=\int R{\Big (}{\frac {dt^{n}-b}{a-ct^{n}}},\;t{\Big )}\;{\frac {(ad-bc)nt^{n-1}}{(a-ct^{n})^{2}}}\;dt.}
Kadangi racionalioji funkcija, kurios argumentas yra racionalioji funkcija, irgi yra racionalioji funkcija, tai integralas, parašytas paskutinės lygybės dešinėje pusėje yra racionaliosios funkcijos integralas. Vadinasi įrodėme, kad (7.64) trupmeninio tiesinio iracionalumo integralas racionalizuojamas keitiniu
t
=
a
x
+
b
c
x
+
d
n
.
{\displaystyle t={\sqrt[{n}]{\frac {ax+b}{cx+d}}}.}
Pavyzdžiai.
Apskaičiuosime integralą
I
=
∫
1
+
x
1
−
x
⋅
d
x
1
−
x
.
{\displaystyle I=\int {\sqrt {\frac {1+x}{1-x}}}\cdot {\frac {dx}{1-x}}.}
t
=
1
+
x
1
−
x
,
t
2
=
1
+
x
1
−
x
,
t
2
−
x
t
2
=
1
+
x
,
−
x
t
2
−
x
=
1
−
t
2
,
x
(
t
2
+
1
)
=
t
2
−
1
,
x
=
t
2
−
1
t
2
+
1
,
{\displaystyle t={\sqrt {\frac {1+x}{1-x}}},\;\;t^{2}={\frac {1+x}{1-x}},\;\;t^{2}-xt^{2}=1+x,\;\;-xt^{2}-x=1-t^{2},\;\;x(t^{2}+1)=t^{2}-1,\;\;x={\frac {t^{2}-1}{t^{2}+1}},}
d
x
=
2
t
(
t
2
+
1
)
−
(
t
2
−
1
)
2
t
(
t
2
+
1
)
2
d
t
=
2
t
3
+
2
t
−
2
t
3
+
2
t
(
t
2
+
1
)
2
d
t
=
4
t
d
t
(
t
2
+
1
)
2
,
{\displaystyle dx={\frac {2t(t^{2}+1)-(t^{2}-1)2t}{(t^{2}+1)^{2}}}\;dt={\frac {2t^{3}+2t-2t^{3}+2t}{(t^{2}+1)^{2}}}\;dt={\frac {4t\;dt}{(t^{2}+1)^{2}}},}
gauname
I
=
∫
1
+
x
1
−
x
⋅
d
x
1
−
x
=
∫
1
+
t
2
−
1
t
2
+
1
1
−
t
2
−
1
t
2
+
1
⋅
4
t
d
t
(
t
2
+
1
)
2
1
−
t
2
−
1
t
2
+
1
=
∫
t
2
+
1
+
t
2
−
1
t
2
+
1
t
2
+
1
−
(
t
2
−
1
)
t
2
+
1
⋅
4
t
d
t
(
t
2
+
1
)
2
t
2
+
1
−
(
t
2
−
1
)
t
2
+
1
=
{\displaystyle I=\int {\sqrt {\frac {1+x}{1-x}}}\cdot {\frac {dx}{1-x}}=\int {\sqrt {\frac {1+{\frac {t^{2}-1}{t^{2}+1}}}{1-{\frac {t^{2}-1}{t^{2}+1}}}}}\cdot {\frac {\frac {4t\;dt}{(t^{2}+1)^{2}}}{1-{\frac {t^{2}-1}{t^{2}+1}}}}=\int {\sqrt {\frac {\frac {t^{2}+1+t^{2}-1}{t^{2}+1}}{\frac {t^{2}+1-(t^{2}-1)}{t^{2}+1}}}}\cdot {\frac {\frac {4t\;dt}{(t^{2}+1)^{2}}}{\frac {t^{2}+1-(t^{2}-1)}{t^{2}+1}}}=}
=
∫
2
t
2
t
2
+
1
2
t
2
+
1
⋅
4
t
d
t
(
t
2
+
1
)
2
2
t
2
+
1
=
∫
2
t
2
2
⋅
2
t
d
t
t
2
+
1
=
2
∫
t
2
d
t
t
2
+
1
=
2
∫
d
t
−
2
∫
d
t
t
2
+
1
=
2
t
−
2
arctan
t
+
C
=
{\displaystyle =\int {\sqrt {\frac {\frac {2t^{2}}{t^{2}+1}}{\frac {2}{t^{2}+1}}}}\cdot {\frac {\frac {4t\;dt}{(t^{2}+1)^{2}}}{\frac {2}{t^{2}+1}}}=\int {\sqrt {\frac {2t^{2}}{2}}}\cdot {\frac {2t\;dt}{t^{2}+1}}=2\int {\frac {t^{2}\;dt}{t^{2}+1}}=2\int dt-2\int {\frac {dt}{t^{2}+1}}=2t-2\arctan t+C=}
=
2
1
+
x
1
−
x
−
2
arctan
1
+
x
1
−
x
+
C
.
{\displaystyle =2{\sqrt {\frac {1+x}{1-x}}}-2\arctan {\sqrt {\frac {1+x}{1-x}}}+C.}
5. Kvadratinių iracionalumų integravimas kitais būdais
keisti
Nors Oilerio keitiniais
R
(
x
,
a
x
2
+
b
x
+
c
)
(
7.66
)
{\displaystyle R(x,\;{\sqrt {ax^{2}+bx+c}})\quad (7.66)}
funkcijos integralas visada racionalizuojamas, bet dažniausiai gaunami labai griozdiški ir sudėtingi reiškiniai. Todėl praktikoje (7.66) funkcija integruojama kitais būdais. Juos ir aptarsime šiame skirsnyje.
Pažymėję
y
=
a
x
2
+
b
x
+
c
{\displaystyle y={\sqrt {ax^{2}+bx+c}}}
y
2
{\displaystyle y^{2}}
R
(
x
,
y
)
=
R
1
(
x
)
+
R
2
(
x
)
y
;
{\displaystyle R(x,\;y)=R_{1}(x)+{\frac {R_{2}(x)}{y}};}
(vietoje
R
2
(
x
)
{\displaystyle R_{2}(x)}
y
2
{\displaystyle y^{2}}
y būtų skaityklyje; taip pat galime parinkti
R
2
(
x
)
,
{\displaystyle R_{2}(x),}
y
2
{\displaystyle y^{2}}
y
2
{\displaystyle y^{2}}
čia
R
1
(
x
)
{\displaystyle R_{1}(x)}
R
2
(
x
)
{\displaystyle R_{2}(x)}
R
1
(
x
)
{\displaystyle R_{1}(x)}
R
2
(
x
)
y
{\displaystyle {\frac {R_{2}(x)}{y}}}
Žinome, kad kiekvieną racionaliąją trupmeną
R
2
(
x
)
{\displaystyle R_{2}(x)}
P (x ) ir taisiklingos racionaliosios trupmenos
R
3
(
x
)
{\displaystyle R_{3}(x)}
R
3
(
x
)
{\displaystyle R_{3}(x)}
R
2
(
x
)
y
{\displaystyle {\frac {R_{2}(x)}{y}}}
I.
∫
P
(
x
)
y
d
x
,
{\displaystyle \int {\frac {P(x)}{y}}\;dx,\;}
II.
∫
B
(
x
−
A
)
α
y
d
x
,
{\displaystyle \int {\frac {B}{(x-A)^{\alpha }y}}\;dx,\;}
A ir B - realieji skaičiai, o
α
{\displaystyle \alpha }
III.
∫
M
x
+
N
(
x
2
+
p
x
+
q
)
λ
y
d
x
,
{\displaystyle \int {\frac {Mx+N}{(x^{2}+px+q)^{\lambda }y}}\;dx,\;}
M, N, p ir q - realieji skaičiai,
λ
{\displaystyle \lambda }
q
−
p
2
4
>
0
{\displaystyle q-{\frac {p^{2}}{4}}>0\;}
x
2
+
p
x
+
q
{\displaystyle x^{2}+px+q}
Visų trijų tipų integralus išnagrinėsime atskirai .Pirmojo tipo integralas
keisti
I. Pirmojo tipo integralui apskaičiuoti pirmiausia išvesime integralo
∫
x
m
d
x
y
(
m
=
0
,
1
,
2
,
.
.
.
)
{\displaystyle \int {\frac {x^{m}\;dx}{y}}\;\;(m=0,\;1,\;2,\;...)}
rekurentinę formulę. Tarę, kad
m
≥
1
,
{\displaystyle m\geq 1,}
(
x
m
−
1
y
)
′
=
(
x
m
−
1
a
x
2
+
b
x
+
c
)
′
=
(
m
−
1
)
x
m
−
2
a
x
2
+
b
x
+
c
+
x
m
−
1
⋅
2
a
x
+
b
2
a
x
2
+
b
x
+
c
=
{\displaystyle (x^{m-1}y)'=(x^{m-1}{\sqrt {ax^{2}+bx+c}})'=(m-1)x^{m-2}{\sqrt {ax^{2}+bx+c}}+x^{m-1}\cdot {\frac {2ax+b}{2{\sqrt {ax^{2}+bx+c}}}}=}
=
(
m
−
1
)
x
m
−
2
a
x
2
+
b
x
+
c
+
a
x
m
a
x
2
+
b
x
+
c
+
x
m
−
1
b
2
a
x
2
+
b
x
+
c
=
(
m
−
1
)
x
m
−
2
a
x
2
+
b
x
+
c
a
x
2
+
b
x
+
c
+
a
x
m
a
x
2
+
b
x
+
c
+
x
m
−
1
b
2
a
x
2
+
b
x
+
c
=
{\displaystyle =(m-1)x^{m-2}{\sqrt {ax^{2}+bx+c}}+{\frac {ax^{m}}{\sqrt {ax^{2}+bx+c}}}+{\frac {x^{m-1}b}{2{\sqrt {ax^{2}+bx+c}}}}=(m-1)x^{m-2}{\frac {ax^{2}+bx+c}{\sqrt {ax^{2}+bx+c}}}+{\frac {ax^{m}}{\sqrt {ax^{2}+bx+c}}}+{\frac {x^{m-1}b}{2{\sqrt {ax^{2}+bx+c}}}}=}
=
(
m
−
1
)
a
x
m
a
x
2
+
b
x
+
c
+
(
m
−
1
)
b
x
m
−
1
a
x
2
+
b
x
+
c
+
(
m
−
1
)
c
x
m
−
2
a
x
2
+
b
x
+
c
+
a
x
m
a
x
2
+
b
x
+
c
+
x
m
−
1
b
2
a
x
2
+
b
x
+
c
=
{\displaystyle =(m-1){\frac {ax^{m}}{\sqrt {ax^{2}+bx+c}}}+(m-1){\frac {bx^{m-1}}{\sqrt {ax^{2}+bx+c}}}+(m-1){\frac {cx^{m-2}}{\sqrt {ax^{2}+bx+c}}}+{\frac {ax^{m}}{\sqrt {ax^{2}+bx+c}}}+{\frac {x^{m-1}b}{2{\sqrt {ax^{2}+bx+c}}}}=}
=
m
a
x
m
a
x
2
+
b
x
+
c
+
(
m
−
1
2
)
b
x
m
−
1
a
x
2
+
b
x
+
c
+
(
m
−
1
)
c
x
m
−
2
a
x
2
+
b
x
+
c
,
{\displaystyle =m{\frac {ax^{m}}{\sqrt {ax^{2}+bx+c}}}+(m-{\frac {1}{2}}){\frac {bx^{m-1}}{\sqrt {ax^{2}+bx+c}}}+(m-1){\frac {cx^{m-2}}{\sqrt {ax^{2}+bx+c}}},}
(
x
m
−
1
y
)
′
=
m
a
x
m
y
+
(
m
−
1
2
)
b
x
m
−
1
y
+
(
m
−
1
)
c
x
m
−
2
y
.
{\displaystyle (x^{m-1}y)'=ma{\frac {x^{m}}{y}}+{\Big (}m-{\frac {1}{2}}{\Big )}b{\frac {x^{m-1}}{y}}+(m-1)c{\frac {x^{m-2}}{y}}.}
Suintegravę šią tapatybę panariui, gauname lygybę
x
m
−
1
y
=
m
a
I
m
+
(
m
−
1
2
)
b
I
m
−
1
+
(
m
−
1
)
c
I
m
−
2
.
(
7.69
)
{\displaystyle x^{m-1}y=maI_{m}+{\Big (}m-{\frac {1}{2}}{\Big )}bI_{m-1}+(m-1)cI_{m-2}.\quad (7.69)}
Į (7.69) lygybę įrašę
m
=
1
,
{\displaystyle m=1,}
x
1
−
1
y
=
1
⋅
a
I
1
+
(
1
−
1
2
)
b
I
1
−
1
+
(
1
−
1
)
c
I
1
−
2
,
{\displaystyle x^{1-1}y=1\cdot aI_{1}+{\Big (}1-{\frac {1}{2}}{\Big )}bI_{1-1}+(1-1)cI_{1-2},}
y
=
a
I
1
+
1
2
b
I
0
,
{\displaystyle y=aI_{1}+{\frac {1}{2}}bI_{0},}
y
−
1
2
b
I
0
=
a
I
1
,
{\displaystyle y-{\frac {1}{2}}bI_{0}=aI_{1},}
I
1
=
y
a
−
b
2
a
I
0
.
(
7.70
)
{\displaystyle I_{1}={\frac {y}{a}}-{\frac {b}{2a}}I_{0}.\quad (7.70)}
Į (7.69) lygybę įrašę
m
=
2
{\displaystyle m=2}
I
1
{\displaystyle I_{1}}
x
2
−
1
y
=
2
a
I
2
+
(
2
−
1
2
)
b
I
2
−
1
+
(
2
−
1
)
c
I
2
−
2
,
{\displaystyle x^{2-1}y=2aI_{2}+{\Big (}2-{\frac {1}{2}}{\Big )}bI_{2-1}+(2-1)cI_{2-2},}
x
y
=
2
a
I
2
+
3
2
b
I
1
+
c
I
0
,
{\displaystyle xy=2aI_{2}+{\frac {3}{2}}bI_{1}+cI_{0},}
x
y
−
3
2
b
I
1
−
c
I
0
=
2
a
I
2
,
{\displaystyle xy-{\frac {3}{2}}bI_{1}-cI_{0}=2aI_{2},}
I
2
=
x
y
2
a
−
3
b
4
a
I
1
−
c
2
a
I
0
,
{\displaystyle I_{2}={\frac {xy}{2a}}-{\frac {3b}{4a}}I_{1}-{\frac {c}{2a}}I_{0},}
I
2
=
x
y
2
a
−
3
b
4
a
(
y
a
−
b
2
a
I
0
)
−
c
2
a
I
0
,
{\displaystyle I_{2}={\frac {xy}{2a}}-{\frac {3b}{4a}}({\frac {y}{a}}-{\frac {b}{2a}}I_{0})-{\frac {c}{2a}}I_{0},}
I
2
=
x
y
2
a
−
3
b
y
4
a
2
+
3
b
2
8
a
2
I
0
−
c
2
a
I
0
,
{\displaystyle I_{2}={\frac {xy}{2a}}-{\frac {3by}{4a^{2}}}+{\frac {3b^{2}}{8a^{2}}}I_{0}-{\frac {c}{2a}}I_{0},}
I
2
=
1
4
a
2
(
2
a
x
−
3
b
)
y
+
1
8
a
2
(
3
b
2
−
4
a
c
)
I
0
.
{\displaystyle I_{2}={\frac {1}{4a^{2}}}(2ax-3b)y+{\frac {1}{8a^{2}}}(3b^{2}-4ac)I_{0}.}
Taip samprotaudami toliau, išvesime bendrąją formulę
I
m
=
P
m
−
1
(
x
)
y
+
c
m
I
0
;
(
7.71
)
{\displaystyle I_{m}=P_{m-1}(x)y+c_{m}I_{0};\quad (7.71)}
čia
P
m
−
1
(
x
)
{\displaystyle P_{m-1}(x)}
c
m
{\displaystyle c_{m}}
Jei I tipo integrale P(x) yra n-ojo laipsnio polinomas, tai I tipo integralas bus lygus integralų
I
0
,
I
1
,
.
.
.
,
I
n
{\displaystyle I_{0},\;I_{1},\;...,\;I_{n}\;}
I tipo integralas galutinai apskaičiuojamas pagal tokią formulę:
∫
P
(
x
)
y
d
x
=
Q
n
−
1
y
+
C
0
∫
d
x
y
;
(
7.72
)
{\displaystyle \int {\frac {P(x)}{y}}\;dx=Q_{n-1}y+C_{0}\int {\frac {dx}{y}};\quad (7.72)}
čia
Q
n
−
1
(
x
)
{\displaystyle Q_{n-1}(x)}
C
0
{\displaystyle C_{0}}
Q
n
−
1
(
x
)
{\displaystyle Q_{n-1}(x)}
C
0
{\displaystyle C_{0}}
neapibrėžtųjų koeficientų metodu . Tam reikalui polinomą
Q
n
−
1
(
x
)
{\displaystyle Q_{n-1}(x)}
Q
n
−
1
(
x
)
=
A
0
+
A
1
x
+
.
.
.
+
A
n
−
1
x
n
−
1
.
{\displaystyle Q_{n-1}(x)=A_{0}+A_{1}x+...+A_{n-1}x^{n-1}.}
Išdiferencijavę (7.72) lygybę ir diferencijavimo rezultatą padauginę iš y , gauname
P
(
x
)
y
=
(
Q
n
−
1
y
)
′
+
C
0
1
y
,
{\displaystyle {\frac {P(x)}{y}}=(Q_{n-1}y)'+C_{0}{\frac {1}{y}},}
P
(
x
)
y
=
Q
n
−
1
′
y
+
Q
n
−
1
2
a
x
+
b
2
y
+
C
0
1
y
,
{\displaystyle {\frac {P(x)}{y}}=Q_{n-1}'y+Q_{n-1}{\frac {2ax+b}{2y}}+C_{0}{\frac {1}{y}},}
P
(
x
)
=
Q
n
−
1
′
(
a
x
2
+
b
x
+
c
)
+
1
2
Q
n
−
1
(
2
a
x
+
b
)
+
C
0
.
(
7.73
)
{\displaystyle P(x)=Q_{n-1}'(ax^{2}+bx+c)+{\frac {1}{2}}Q_{n-1}(2ax+b)+C_{0}.\quad (7.73)}
Abiejose tos lygybės pusėse parašyti n -ojo laipsnio polinomai. Palyginę jų koeficientus, sudarome
n
+
1
{\displaystyle n+1}
A
0
,
A
1
,
.
.
.
,
A
n
−
1
{\displaystyle A_{0},\;A_{1},\;...,A_{n-1}\;}
C
0
.
{\displaystyle C_{0}.}
t
=
x
+
b
2
a
{\displaystyle \;t=x+{\frac {b}{2a}}\;}
∫
d
x
y
,
{\displaystyle \int {\frac {dx}{y}},}
∫
d
t
t
2
±
k
2
=
ln
|
t
+
t
2
±
k
2
|
+
C
{\displaystyle \int {\frac {dt}{\sqrt {t^{2}\pm k^{2}}}}=\ln |t+{\sqrt {t^{2}\pm k^{2}}}|+C\;\;}
k
=
const
>
0
{\displaystyle k={\text{const}}>0}
arba integralu
∫
d
t
k
2
−
t
2
=
arcsin
t
k
+
C
.
{\displaystyle \int {\frac {dt}{\sqrt {k^{2}-t^{2}}}}=\arcsin {\frac {t}{k}}+C.}
Pavyzdys . Apskaičiuosime integralą
∫
x
3
1
+
2
x
−
x
2
d
x
.
{\displaystyle \int {\frac {x^{3}}{\sqrt {1+2x-x^{2}}}}\;dx.}
(7.72) formulė, pritaikyta pateiktajam integralui, bus šitokia:
∫
x
3
1
+
2
x
−
x
2
d
x
=
(
A
0
+
A
1
x
+
A
2
x
2
)
1
+
2
x
−
x
2
+
C
0
∫
d
x
1
+
2
x
−
x
2
.
(
7.74
)
{\displaystyle \int {\frac {x^{3}}{\sqrt {1+2x-x^{2}}}}\;dx=(A_{0}+A_{1}x+A_{2}x^{2}){\sqrt {1+2x-x^{2}}}+C_{0}\int {\frac {dx}{\sqrt {1+2x-x^{2}}}}.\quad (7.74)}
Išdiferencijavę abi šios lygybės puses ir rezultatą padauginę iš
1
+
2
x
−
x
2
,
{\displaystyle {\sqrt {1+2x-x^{2}}},}
x
3
1
+
2
x
−
x
2
=
(
A
1
+
2
A
2
x
)
1
+
2
x
−
x
2
+
(
A
0
+
A
1
x
+
A
2
x
2
)
2
−
2
x
2
1
+
2
x
−
x
2
+
C
0
1
1
+
2
x
−
x
2
,
{\displaystyle {\frac {x^{3}}{\sqrt {1+2x-x^{2}}}}=(A_{1}+2A_{2}x){\sqrt {1+2x-x^{2}}}+(A_{0}+A_{1}x+A_{2}x^{2}){\frac {2-2x}{2{\sqrt {1+2x-x^{2}}}}}+C_{0}{\frac {1}{\sqrt {1+2x-x^{2}}}},}
x
3
=
(
A
1
+
2
A
2
x
)
(
1
+
2
x
−
x
2
)
+
(
A
0
+
A
1
x
+
A
2
x
2
)
(
1
−
x
)
+
C
0
.
{\displaystyle x^{3}=(A_{1}+2A_{2}x)(1+2x-x^{2})+(A_{0}+A_{1}x+A_{2}x^{2})(1-x)+C_{0}.}
Palyginę koeficientus prie
x
3
,
x
2
,
x
1
{\displaystyle x^{3},\;x^{2},\;x^{1}\;}
x
0
{\displaystyle x^{0}}
{
−
3
A
2
=
1
,
5
A
2
−
2
A
1
=
0
,
2
A
2
+
3
A
1
−
A
0
=
0
,
A
1
+
A
0
+
C
0
=
0.
{\displaystyle {\begin{cases}-3A_{2}=1,&\\5A_{2}-2A_{1}=0,&\\2A_{2}+3A_{1}-A_{0}=0,&\\A_{1}+A_{0}+C_{0}=0.&\end{cases}}}
Išsprendę šią sistemą, randame
A
2
=
−
1
3
,
A
1
=
−
5
6
,
A
0
=
−
19
6
,
C
0
=
4.
{\displaystyle A_{2}=-{\frac {1}{3}},\;A_{1}=-{\frac {5}{6}},\;A_{0}=-{\frac {19}{6}},\;C_{0}=4.}
t
=
x
−
1
{\displaystyle t=x-1}
∫
d
x
1
+
2
x
−
x
2
=
∫
d
t
2
−
t
2
=
arcsin
t
2
+
C
=
arcsin
x
−
1
2
+
C
.
{\displaystyle \int {\frac {dx}{\sqrt {1+2x-x^{2}}}}=\int {\frac {dt}{\sqrt {2-t^{2}}}}=\arcsin {\frac {t}{\sqrt {2}}}+C=\arcsin {\frac {x-1}{\sqrt {2}}}+C.}
Galutinai
∫
x
3
1
+
2
x
−
x
2
d
x
=
(
−
19
6
−
5
6
x
−
1
3
x
2
)
1
+
2
x
−
x
2
+
4
arcsin
x
−
1
2
+
C
.
{\displaystyle \int {\frac {x^{3}}{\sqrt {1+2x-x^{2}}}}\;dx={\Big (}-{\frac {19}{6}}-{\frac {5}{6}}x-{\frac {1}{3}}x^{2}{\Big )}{\sqrt {1+2x-x^{2}}}+4\arcsin {\frac {x-1}{\sqrt {2}}}+C.}
Antrojo tipo integralas
keisti
II. Skaičiuosime II tipo integralą . Įsitikinsime, kad iš jo keitiniu
t
=
1
x
−
A
{\displaystyle t={\frac {1}{x-A}}}
I tipo integralas. Iš tikrųjų, kadangi
x
−
A
=
1
t
,
x
=
1
t
+
A
,
{\displaystyle x-A={\frac {1}{t}},\;\;x={\frac {1}{t}}+A,\;}
d
x
=
−
d
t
t
2
,
a
x
2
+
b
x
+
c
=
(
A
2
a
+
A
b
+
c
)
t
2
+
(
2
a
A
+
b
)
t
+
a
t
2
,
{\displaystyle dx=-{\frac {dt}{t^{2}}},\;\;ax^{2}+bx+c={\frac {(A^{2}a+Ab+c)t^{2}+(2aA+b)t+a}{t^{2}}},}
tai
∫
B
(
x
−
A
)
α
y
d
x
=
∫
B
(
1
t
)
α
(
A
2
a
+
A
b
+
c
)
t
2
+
(
2
a
A
+
b
)
t
+
a
t
2
⋅
−
d
t
t
2
=
−
∫
B
t
α
(
A
2
a
+
A
b
+
c
)
t
2
+
(
2
a
A
+
b
)
t
+
a
t
⋅
d
t
t
2
=
{\displaystyle \int {\frac {B}{(x-A)^{\alpha }y}}\;dx=\int {\frac {B}{({\frac {1}{t}})^{\alpha }{\sqrt {\frac {(A^{2}a+Ab+c)t^{2}+(2aA+b)t+a}{t^{2}}}}}}\cdot {\frac {-dt}{t^{2}}}=-\int {\frac {Bt^{\alpha }}{\frac {\sqrt {(A^{2}a+Ab+c)t^{2}+(2aA+b)t+a}}{t}}}\cdot {\frac {dt}{t^{2}}}=}
=
−
∫
B
t
α
+
1
(
A
2
a
+
A
b
+
c
)
t
2
+
(
2
a
A
+
b
)
t
+
a
⋅
d
t
t
2
=
−
∫
B
t
α
−
1
(
A
2
a
+
A
b
+
c
)
t
2
+
(
2
a
A
+
b
)
t
+
a
d
t
.
{\displaystyle =-\int {\frac {Bt^{\alpha +1}}{\sqrt {(A^{2}a+Ab+c)t^{2}+(2aA+b)t+a}}}\cdot {\frac {dt}{t^{2}}}=-\int {\frac {Bt^{\alpha -1}}{\sqrt {(A^{2}a+Ab+c)t^{2}+(2aA+b)t+a}}}\;dt.}
Trečiojo tipo integralas
keisti
III. Aptarsime pagaliau, kaip skaičiuojamas III tipo integralas . Pirmiausia išnagrinėsime, kaip skaičiuoti III tipo integralą tuo atveju, kai
p
=
0
,
b
=
0
,
{\displaystyle p=0,\;b=0,}
K
=
∫
M
x
+
N
(
x
2
+
q
)
λ
a
x
2
+
c
d
x
.
{\displaystyle K=\int {\frac {Mx+N}{(x^{2}+q)^{\lambda }{\sqrt {ax^{2}+c}}}}\;dx.}
Jis yra dviejų integralų
K
1
=
M
∫
x
d
x
(
x
2
+
q
)
λ
a
x
2
+
c
{\displaystyle K_{1}=M\int {\frac {x\;dx}{(x^{2}+q)^{\lambda }{\sqrt {ax^{2}+c}}}}\;}
K
2
=
N
∫
d
x
(
x
2
+
q
)
λ
a
x
2
+
c
{\displaystyle \;K_{2}=N\int {\frac {dx}{(x^{2}+q)^{\lambda }{\sqrt {ax^{2}+c}}}}}
suma. Pirmąjį iš jų galima užrašyti šitaip:
K
1
=
M
2
∫
d
(
x
2
)
(
x
2
+
q
)
λ
a
x
2
+
c
;
{\displaystyle K_{1}={\frac {M}{2}}\int {\frac {d(x^{2})}{(x^{2}+q)^{\lambda }{\sqrt {ax^{2}+c}}}}\;;}
iš čia aišku, kad pointegralinė funkcija yra tiesinis (bet ne kvadratinis) iracionalumas
x
2
{\displaystyle x^{2}}
K
1
{\displaystyle K_{1}}
t
=
a
x
2
+
c
.
{\displaystyle t={\sqrt {ax^{2}+c}}.}
K
2
{\displaystyle K_{2}}
x
≠
0
{\displaystyle x\neq 0}
K
2
=
N
∫
d
x
(
x
2
+
q
)
λ
a
x
2
+
c
=
N
∫
d
x
x
2
λ
(
1
+
q
1
x
2
)
λ
x
a
+
c
1
x
2
=
N
∫
1
x
2
λ
+
1
d
x
(
1
+
q
1
x
2
)
λ
a
+
c
1
x
2
=
{\displaystyle K_{2}=N\int {\frac {dx}{(x^{2}+q)^{\lambda }{\sqrt {ax^{2}+c}}}}=N\int {\frac {dx}{x^{2\lambda }{\big (}1+q{\frac {1}{x^{2}}}{\big )}^{\lambda }x{\sqrt {a+c{\frac {1}{x^{2}}}}}}}=N\int {\frac {{\frac {1}{x^{2\lambda +1}}}dx}{{\big (}1+q{\frac {1}{x^{2}}}{\big )}^{\lambda }{\sqrt {a+c{\frac {1}{x^{2}}}}}}}=}
=
N
∫
1
x
2
λ
−
2
d
x
x
3
(
1
+
q
1
x
2
)
λ
a
+
c
1
x
2
=
−
N
2
∫
(
1
x
2
)
λ
−
1
d
(
1
x
2
)
(
1
+
q
1
x
2
)
λ
a
+
c
1
x
2
,
{\displaystyle =N\int {\frac {{\frac {1}{x^{2\lambda -2}}}\;{\frac {dx}{x^{3}}}}{{\big (}1+q{\frac {1}{x^{2}}}{\big )}^{\lambda }{\sqrt {a+c{\frac {1}{x^{2}}}}}}}=-{\frac {N}{2}}\int {\frac {{\big (}{\frac {1}{x^{2}}}{\big )}^{\lambda -1}\;d{\big (}{\frac {1}{x^{2}}}{\big )}}{{\big (}1+q{\frac {1}{x^{2}}}{\big )}^{\lambda }{\sqrt {a+c{\frac {1}{x^{2}}}}}}}\;,}
o iš to aišku, kad pointegralinė funkcija yra tiesinis iracionalumas
1
x
2
{\displaystyle {\frac {1}{x^{2}}}}
K
2
{\displaystyle K_{2}}
r
=
a
+
c
x
2
.
{\displaystyle r={\sqrt {a+{\frac {c}{x^{2}}}}}.}
III tipo integralą racionalizavome atskiru atveju, kai abu kvadratiniai trinariai neturi pirmojo laipsnio narių .
Dabar išnagrinėsime III tipo integralą bendruoju atveju ir įrodysime, kad jį galima pakeisti ką tik ištirtu integralu. Jei kvadratinių trinarių koeficientai yra susieti lygybe
b
=
a
p
,
(
7.75
)
{\displaystyle b=ap,\quad (7.75)}
tai III tipo integralo pakeitimui jau ištirtuoju integralu pakanka keitinio
x
=
t
−
p
2
.
{\displaystyle x=t-{\frac {p}{2}}.}
∫
(
M
x
+
N
)
d
x
(
x
2
+
p
x
+
q
)
λ
a
x
2
+
b
x
+
c
=
∫
M
t
+
(
−
M
p
2
+
N
)
[
t
2
+
(
q
−
p
2
4
)
]
λ
a
t
2
+
(
c
−
a
p
2
4
)
d
t
.
{\displaystyle \int {\frac {(Mx+N)\;dx}{(x^{2}+px+q)^{\lambda }{\sqrt {ax^{2}+bx+c}}}}=\int {\frac {Mt+{\Big (}{\frac {-Mp}{2}}+N{\Big )}}{{\Big [}t^{2}+{\Big (}q-{\frac {p^{2}}{4}}{\Big )}{\Big ]}^{\lambda }{\sqrt {at^{2}+{\Big (}c-{\frac {ap^{2}}{4}}{\Big )}}}}}\;dt.}
(Iš tikro, vietoje t dešinėje lygybės pusėje įstatę
x
+
p
2
,
{\displaystyle x+{\frac {p}{2}},}
∫
M
t
+
(
−
M
p
2
+
N
)
[
t
2
+
(
q
−
p
2
4
)
]
λ
a
t
2
+
(
c
−
a
p
2
4
)
d
t
=
∫
M
(
x
+
p
2
)
+
(
−
M
p
2
+
N
)
[
(
x
+
p
2
)
2
+
(
q
−
p
2
4
)
]
λ
a
(
x
+
p
2
)
2
+
(
c
−
a
p
2
4
)
d
x
=
{\displaystyle \int {\frac {Mt+{\Big (}{\frac {-Mp}{2}}+N{\Big )}}{{\Big [}t^{2}+{\Big (}q-{\frac {p^{2}}{4}}{\Big )}{\Big ]}^{\lambda }{\sqrt {at^{2}+{\Big (}c-{\frac {ap^{2}}{4}}{\Big )}}}}}\;dt=\int {\frac {M(x+{\frac {p}{2}})+{\Big (}{\frac {-Mp}{2}}+N{\Big )}}{{\Big [}(x+{\frac {p}{2}})^{2}+{\Big (}q-{\frac {p^{2}}{4}}{\Big )}{\Big ]}^{\lambda }{\sqrt {a(x+{\frac {p}{2}})^{2}+{\Big (}c-{\frac {ap^{2}}{4}}{\Big )}}}}}\;dx=}
=
∫
M
x
+
M
p
2
+
−
M
p
2
+
N
[
x
2
+
p
x
+
p
2
4
+
(
q
−
p
2
4
)
]
λ
a
(
x
2
+
p
x
+
p
2
4
)
+
(
c
−
a
p
2
4
)
d
x
=
∫
M
x
+
N
[
x
2
+
p
x
+
q
]
λ
a
x
2
+
a
p
x
+
a
p
2
4
+
c
−
a
p
2
4
d
x
=
{\displaystyle =\int {\frac {Mx+{\frac {Mp}{2}}+{\frac {-Mp}{2}}+N}{{\Big [}x^{2}+px+{\frac {p^{2}}{4}}+{\Big (}q-{\frac {p^{2}}{4}}{\Big )}{\Big ]}^{\lambda }{\sqrt {a(x^{2}+px+{\frac {p^{2}}{4}})+{\Big (}c-{\frac {ap^{2}}{4}}{\Big )}}}}}\;dx=\int {\frac {Mx+N}{{\Big [}x^{2}+px+q{\Big ]}^{\lambda }{\sqrt {ax^{2}+apx+{\frac {ap^{2}}{4}}+c-{\frac {ap^{2}}{4}}}}}}\;dx=}
=
∫
M
x
+
N
[
x
2
+
p
x
+
q
]
λ
a
x
2
+
a
p
x
+
c
d
x
=
∫
(
M
x
+
N
)
d
x
(
x
2
+
p
x
+
q
)
λ
a
x
2
+
b
x
+
c
.
{\displaystyle =\int {\frac {Mx+N}{[x^{2}+px+q]^{\lambda }{\sqrt {ax^{2}+apx+c}}}}\;dx=\int {\frac {(Mx+N)\;dx}{(x^{2}+px+q)^{\lambda }{\sqrt {ax^{2}+bx+c}}}}.}
Nes
b
=
a
p
{\displaystyle b=ap\;}
d
x
=
d
(
t
−
p
2
)
=
d
t
.
{\displaystyle \;dx=d{\Big (}t-{\frac {p}{2}}{\Big )}=dt.}
III tipo integralą sunkiau pakeisti anksčiau ištirtuoju integralu tuo atveju, kai kvadratinių trinarių koeficientai nėra susieti (7.75) lygybe. Tuomet iš pradžių reikia naudotis trupmeniniu tiesiniu keitiniu
x
=
μ
t
+
ν
1
+
t
,
(
7.76
)
{\displaystyle x={\frac {\mu t+\nu }{1+t}}\;,\quad (7.76)}
taip pasirinkus koeficientus
μ
{\displaystyle \mu }
ν
,
{\displaystyle \nu ,}
neturėtų pirmojo laipsnio narių t atžvilgiu . Įsitikinsime, kad tokius
μ
{\displaystyle \mu }
ν
{\displaystyle \nu }
x
2
+
p
x
+
q
=
(
μ
2
+
p
μ
+
q
)
t
2
+
[
2
μ
ν
+
p
(
μ
+
ν
)
+
2
q
]
t
+
(
ν
2
+
p
ν
+
q
)
(
1
+
t
)
2
,
{\displaystyle x^{2}+px+q={\frac {(\mu ^{2}+p\mu +q)t^{2}+[2\mu \nu +p(\mu +\nu )+2q]t+(\nu ^{2}+p\nu +q)}{(1+t)^{2}}}\;,}
a
x
2
+
b
x
+
c
=
(
a
μ
2
+
b
μ
+
c
)
t
2
+
[
2
μ
ν
a
+
b
(
μ
+
ν
)
+
2
c
]
t
+
(
a
ν
2
+
b
ν
+
c
)
(
1
+
t
)
2
.
{\displaystyle ax^{2}+bx+c={\frac {(a\mu ^{2}+b\mu +c)t^{2}+[2\mu \nu a+b(\mu +\nu )+2c]t+(a\nu ^{2}+b\nu +c)}{(1+t)^{2}}}\;.}
Vadinasi, koeficientai
μ
{\displaystyle \mu }
ν
{\displaystyle \nu }
{
2
μ
ν
+
p
(
μ
+
ν
)
+
2
q
=
0
,
2
μ
ν
a
+
b
(
μ
+
ν
)
+
2
c
=
0
,
{\displaystyle \quad {\begin{cases}2\mu \nu +p(\mu +\nu )+2q=0,&\\2\mu \nu a+b(\mu +\nu )+2c=0,&\end{cases}}}
arba iš jai ekvivalenčios lygčių sistemos
{
μ
+
ν
=
−
2
(
c
−
a
q
)
b
−
a
p
,
μ
⋅
ν
=
c
p
−
b
q
b
−
a
p
.
{\displaystyle \quad {\begin{cases}\mu +\nu =-{\frac {2(c-aq)}{b-ap}},&\\\mu \cdot \nu ={\frac {cp-bq}{b-ap}}.&\end{cases}}}
(Antroje sistemoje
μ
+
ν
{\displaystyle \mu +\nu }
a ) ir sudėję su antra eilute. Tada gaunasi:
−
a
p
(
μ
+
ν
)
+
b
(
μ
+
ν
)
−
2
a
q
+
2
c
=
0
,
{\displaystyle -ap(\mu +\nu )+b(\mu +\nu )-2aq+2c=0,}
(
μ
+
ν
)
(
b
−
a
p
)
=
2
a
q
−
2
c
,
{\displaystyle (\mu +\nu )(b-ap)=2aq-2c,}
μ
+
ν
=
−
2
(
c
−
a
q
)
b
−
a
p
.
{\displaystyle \mu +\nu =-{\frac {2(c-aq)}{b-ap}}.}
Antroje sistemoje
μ
⋅
ν
{\displaystyle \mu \cdot \nu }
b ), o antrą eilutę padauginę iš p ir abi [pirmos sistemos] eilutes sudėję. Po sudeties turime:
−
2
μ
ν
b
+
2
μ
ν
a
p
−
2
q
b
+
2
c
p
=
0
,
{\displaystyle -2\mu \nu b+2\mu \nu ap-2qb+2cp=0,}
−
μ
ν
b
+
μ
ν
a
p
−
q
b
+
c
p
=
0
,
{\displaystyle -\mu \nu b+\mu \nu ap-qb+cp=0,}
(
a
p
−
b
)
μ
ν
=
q
b
−
c
p
,
{\displaystyle (ap-b)\mu \nu =qb-cp,}
μ
⋅
ν
=
c
p
−
b
q
b
−
a
p
.
{\displaystyle \mu \cdot \nu ={\frac {cp-bq}{b-ap}}.}
Taigi
μ
{\displaystyle \mu }
ν
{\displaystyle \nu }
z
2
+
2
(
c
−
a
q
)
b
−
a
p
z
+
c
p
−
b
q
b
−
a
p
=
0.
(
7.77
)
{\displaystyle z^{2}+{\frac {2(c-aq)}{b-ap}}\;z+{\frac {cp-bq}{b-ap}}=0.\quad (7.77)}
Liko įrodyti, kad ta kvadratinė lygtis turi realias ir skirtingas šaknis (
μ
{\displaystyle \mu }
ν
{\displaystyle \nu }
(
2
(
c
−
a
q
)
b
−
a
p
)
2
>
4
c
p
−
b
q
b
−
a
p
,
{\displaystyle {\Big (}{\frac {2(c-aq)}{b-ap}}{\Big )}^{2}>4{\frac {cp-bq}{b-ap}},}
(
c
−
a
q
)
2
>
(
c
p
−
b
q
)
(
b
−
a
p
)
.
(
7.78
)
{\displaystyle (c-aq)^{2}>(cp-bq)(b-ap).\quad (7.78)}
Lengva patikrinti, kad (7.78) nelygybė yra ekvivalenti nelygybei
[
2
(
c
+
a
q
)
−
b
p
]
2
>
(
4
q
−
p
2
)
(
4
a
c
−
b
2
)
.
(
7.79
)
{\displaystyle [2(c+aq)-bp]^{2}>(4q-p^{2})(4ac-b^{2}).\quad (7.79)}
Kadangi kvadratinis trinaris
x
2
+
p
x
+
q
{\displaystyle x^{2}+px+q}
4
q
−
p
2
>
0.
{\displaystyle 4q-p^{2}>0.}
(7.79) nelygybė tikrai yra teisinga, kai
4
a
c
−
b
2
<
0.
{\displaystyle 4ac-b^{2}<0.}
4
a
c
−
b
2
>
0.
{\displaystyle 4ac-b^{2}>0.}
q
>
0
,
a
c
>
0
{\displaystyle q>0,\;ac>0}
4
a
c
q
>
p
b
{\displaystyle 4{\sqrt {acq}}>pb\;}
4
q
⋅
4
a
c
>
p
2
⋅
b
2
{\displaystyle 4q\cdot 4ac>p^{2}\cdot b^{2}}
c
+
a
q
2
≥
a
c
q
,
{\displaystyle {\frac {c+aq}{2}}\geq {\sqrt {acq}},}
[
2
(
c
+
a
q
)
−
b
p
]
2
≥
[
4
q
a
c
−
b
p
]
2
=
(
4
q
−
p
2
)
(
4
a
c
−
b
2
)
+
4
(
p
a
c
−
b
q
)
2
≥
(
4
q
−
p
2
)
(
4
a
c
−
b
2
)
.
{\displaystyle [2(c+aq)-bp]^{2}\geq [4{\sqrt {qac}}-bp]^{2}=(4q-p^{2})(4ac-b^{2})+4(p{\sqrt {ac}}-b{\sqrt {q}})^{2}\geq (4q-p^{2})(4ac-b^{2}).}
Parašytoje nelygybių grandinėje yra bent vienas griežtos nelygybės ženklas >, nes pirmasis ženklas
≥
{\displaystyle \geq }
=
{\displaystyle =}
c
=
a
q
;
{\displaystyle c=aq;}
c
=
a
q
,
{\displaystyle c=aq,}
b
≠
a
p
{\displaystyle b\neq ap}
p
a
c
−
b
q
≠
0
,
{\displaystyle p{\sqrt {ac}}-b{\sqrt {q}}\neq 0,\;}
≥
{\displaystyle \geq }
μ
{\displaystyle \mu }
ν
,
{\displaystyle \nu ,}
t
{\displaystyle t}
μ
{\displaystyle \mu }
ν
,
{\displaystyle \nu ,}
III tipo integralas pakeičiamas integralu
∫
P
(
t
)
d
t
(
t
2
+
q
1
)
λ
a
1
t
2
+
c
1
;
(
7.80
)
{\displaystyle \int {\frac {P(t)\;dt}{(t^{2}+q_{1})^{\lambda }{\sqrt {a_{1}t^{2}+c_{1}}}}}\;;\quad (7.80)}
čia
a
1
,
c
1
{\displaystyle a_{1},\;c_{1}}
q
1
{\displaystyle q_{1}}
(
2
λ
−
1
)
{\displaystyle (2\lambda -1)}
λ
>
1
{\displaystyle \lambda >1}
P
(
t
)
d
t
(
t
2
+
q
1
)
λ
{\displaystyle \;{\frac {P(t)\;dt}{(t^{2}+q_{1})^{\lambda }}}}
∫
M
k
t
+
N
k
(
t
2
+
q
1
)
k
a
1
t
2
+
c
1
d
t
(
k
=
1
,
2
,
.
.
.
,
λ
)
.
{\displaystyle \int {\frac {M_{k}t+N_{k}}{(t^{2}+q_{1})^{k}{\sqrt {a_{1}t^{2}+c_{1}}}}}\;dt\quad (k=1,\;2,\;...,\;\lambda ).}
Tokie integralai buvo anksčiau ištirti. Įrodėme, kad visų trijų I , II ir III tipų integralai yra elementariosios funkcijos. Vadinasi, dar kartą (be Oilerio keitinių) įsitikinome, kad (7.66) funkcijos integralas išreiškiamas elementariosiomis funkcijomis.
Pavyzdys. Apskaičiuosime integralą
I
=
∫
d
x
(
x
2
−
x
+
1
)
x
2
+
x
+
1
.
{\displaystyle I=\int {\frac {dx}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}.}
III tipo integralas. Kadangi jis netenkina (7.75) sąlygos, tai pirmiausia turime padaryti (7.76) keitinį. Po tokio pakeitimo
x
2
+
x
+
1
=
(
μ
2
+
μ
+
1
)
t
2
+
[
2
μ
ν
+
(
μ
+
ν
)
+
2
]
t
+
(
ν
2
+
ν
+
1
)
(
1
+
t
)
2
,
{\displaystyle x^{2}+x+1={\frac {(\mu ^{2}+\mu +1)t^{2}+[2\mu \nu +(\mu +\nu )+2]t+(\nu ^{2}+\nu +1)}{(1+t)^{2}}}\;,}
x
2
−
x
+
1
=
(
μ
2
−
μ
+
1
)
t
2
+
[
2
μ
ν
−
(
μ
+
ν
)
+
2
]
t
+
(
ν
2
−
ν
+
1
)
(
1
+
t
)
2
.
{\displaystyle x^{2}-x+1={\frac {(\mu ^{2}-\mu +1)t^{2}+[2\mu \nu -(\mu +\nu )+2]t+(\nu ^{2}-\nu +1)}{(1+t)^{2}}}\;.}
Koeficientus
μ
{\displaystyle \mu }
ν
{\displaystyle \nu }
{
2
μ
ν
+
(
μ
+
ν
)
+
2
=
0
,
2
μ
ν
−
(
μ
+
ν
)
+
2
=
0.
{\displaystyle {\begin{cases}2\mu \nu +(\mu +\nu )+2=0,&\\2\mu \nu -(\mu +\nu )+2=0.&\end{cases}}}
Lengva įsitikinti, kad
μ
=
1
,
ν
=
−
1
{\displaystyle \mu =1,\;\nu =-1\;}
μ
=
−
1
,
ν
=
1
{\displaystyle \mu =-1,\;\nu =1}
x
=
t
−
1
t
+
1
,
{\displaystyle x={\frac {t-1}{t+1}}\;,}
x
(
t
+
1
)
=
t
−
1
,
x
t
−
t
=
−
x
−
1
,
t
(
1
−
x
)
=
1
+
x
,
{\displaystyle x(t+1)=t-1,\;xt-t=-x-1,\;t(1-x)=1+x,\;}
t
=
x
+
1
1
−
x
,
{\displaystyle t={\frac {x+1}{1-x}}\;,}
d
x
=
t
+
1
−
(
t
−
1
)
(
t
+
1
)
2
d
t
=
2
d
t
(
t
+
1
)
2
,
{\displaystyle dx={\frac {t+1-(t-1)}{(t+1)^{2}}}\;dt={\frac {2\;dt}{(t+1)^{2}}}\;,}
x
2
+
x
+
1
=
(
t
−
1
)
2
(
t
+
1
)
2
+
t
−
1
t
+
1
+
1
=
t
2
−
2
t
+
1
(
t
+
1
)
2
+
t
−
1
t
+
1
+
1
=
(
t
2
−
2
t
+
1
)
+
(
t
+
1
)
(
t
−
1
)
+
(
t
+
1
)
2
(
t
+
1
)
2
=
{\displaystyle x^{2}+x+1={\frac {(t-1)^{2}}{(t+1)^{2}}}+{\frac {t-1}{t+1}}+1={\frac {t^{2}-2t+1}{(t+1)^{2}}}+{\frac {t-1}{t+1}}+1={\frac {(t^{2}-2t+1)+(t+1)(t-1)+(t+1)^{2}}{(t+1)^{2}}}=}
=
(
t
2
−
2
t
+
1
)
+
(
t
2
−
1
)
+
(
t
2
+
2
t
+
1
)
(
t
+
1
)
2
=
3
t
2
+
1
(
t
+
1
)
2
,
{\displaystyle ={\frac {(t^{2}-2t+1)+(t^{2}-1)+(t^{2}+2t+1)}{(t+1)^{2}}}={\frac {3t^{2}+1}{(t+1)^{2}}}\;,}
x
2
−
x
+
1
=
(
t
2
−
2
t
+
1
)
−
(
t
2
−
1
)
+
(
t
2
+
2
t
+
1
)
(
t
+
1
)
2
=
t
2
+
3
(
t
+
1
)
2
.
{\displaystyle x^{2}-x+1={\frac {(t^{2}-2t+1)-(t^{2}-1)+(t^{2}+2t+1)}{(t+1)^{2}}}={\frac {t^{2}+3}{(t+1)^{2}}}\;.}
Nagrinėjamasis integralas virsta šitokiu:
I
=
∫
d
x
(
x
2
−
x
+
1
)
x
2
+
x
+
1
=
∫
1
t
2
+
3
(
t
+
1
)
2
3
t
2
+
1
(
t
+
1
)
2
⋅
2
d
t
(
t
+
1
)
2
=
2
∫
1
(
t
2
+
3
)
3
t
2
+
1
(
t
+
1
)
2
d
t
=
2
∫
(
t
+
1
)
d
t
(
t
2
+
3
)
3
t
2
+
1
=
I
1
+
I
2
,
{\displaystyle I=\int {\frac {dx}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}=\int {\frac {1}{{\frac {t^{2}+3}{(t+1)^{2}}}{\sqrt {\frac {3t^{2}+1}{(t+1)^{2}}}}}}\cdot {\frac {2\;dt}{(t+1)^{2}}}=2\int {\frac {1}{(t^{2}+3){\sqrt {\frac {3t^{2}+1}{(t+1)^{2}}}}}}\;dt=2\int {\frac {(t+1)dt}{(t^{2}+3){\sqrt {3t^{2}+1}}}}=I_{1}+I_{2},}
jei
I
1
=
2
∫
t
d
t
(
t
2
+
3
)
3
t
2
+
1
,
I
2
=
2
∫
d
t
(
t
2
+
3
)
3
t
2
+
1
.
{\displaystyle I_{1}=2\int {\frac {t\;dt}{(t^{2}+3){\sqrt {3t^{2}+1}}}}\;,\quad I_{2}=2\int {\frac {dt}{(t^{2}+3){\sqrt {3t^{2}+1}}}}.}
Integralui
I
1
{\displaystyle I_{1}}
u
=
3
t
2
+
1
,
{\displaystyle u={\sqrt {3t^{2}+1}},\;}
I
2
{\displaystyle I_{2}}
v
=
3
+
1
t
2
.
{\displaystyle v={\sqrt {3+{\frac {1}{t^{2}}}}}.}
Tada integralui
I
1
{\displaystyle I_{1}}
u
2
=
3
t
2
+
1
,
u
2
−
1
=
3
t
2
,
t
2
=
u
2
−
1
3
,
{\displaystyle u^{2}=3t^{2}+1,\;u^{2}-1=3t^{2},\;t^{2}={\frac {u^{2}-1}{3}}\;,}
d
(
t
2
)
=
d
(
u
2
−
1
3
)
=
2
u
3
d
u
{\displaystyle d(t^{2})=d{\Big (}{\frac {u^{2}-1}{3}}{\Big )}={\frac {2u}{3}}\;du\;}
I
1
=
2
∫
t
d
t
(
t
2
+
3
)
3
t
2
+
1
=
2
2
∫
d
(
t
2
)
(
t
2
+
3
)
3
t
2
+
1
=
∫
2
u
3
d
u
(
u
2
−
1
3
+
3
)
⋅
u
=
2
3
∫
d
u
u
2
−
1
3
+
3
=
{\displaystyle I_{1}=2\int {\frac {t\;dt}{(t^{2}+3){\sqrt {3t^{2}+1}}}}={\frac {2}{2}}\int {\frac {d(t^{2})}{(t^{2}+3){\sqrt {3t^{2}+1}}}}=\int {\frac {{\frac {2u}{3}}\;du}{({\frac {u^{2}-1}{3}}+3)\cdot u}}={\frac {2}{3}}\int {\frac {du}{{\frac {u^{2}-1}{3}}+3}}=}
=
2
3
∫
d
u
u
2
−
1
+
9
3
=
2
∫
d
u
u
2
+
8
=
2
8
arctan
u
8
+
C
=
1
2
arctan
u
8
+
C
=
1
2
arctan
3
t
2
+
1
8
+
C
=
{\displaystyle ={\frac {2}{3}}\int {\frac {du}{\frac {u^{2}-1+9}{3}}}=2\int {\frac {du}{u^{2}+8}}={\frac {2}{\sqrt {8}}}\arctan {\frac {u}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {u}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {3t^{2}+1}}{\sqrt {8}}}+C=}
=
1
2
arctan
3
(
x
+
1
1
−
x
)
2
+
1
8
+
C
=
1
2
arctan
3
(
x
2
+
2
x
+
1
)
(
1
−
x
)
2
+
1
8
+
C
=
1
2
arctan
3
(
x
2
+
2
x
+
1
)
+
1
−
2
x
+
x
2
(
1
−
x
)
2
8
+
C
=
{\displaystyle ={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {3({\frac {x+1}{1-x}})^{2}+1}}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {{\frac {3(x^{2}+2x+1)}{(1-x)^{2}}}+1}}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {\frac {3(x^{2}+2x+1)+1-2x+x^{2}}{(1-x)^{2}}}}{\sqrt {8}}}+C=}
=
1
2
arctan
4
x
2
+
4
x
+
4
8
(
1
−
x
)
2
+
C
=
1
2
arctan
x
2
+
x
+
1
2
(
1
−
x
)
2
+
C
.
{\displaystyle ={\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {4x^{2}+4x+4}{8(1-x)^{2}}}}+C={\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {x^{2}+x+1}{2(1-x)^{2}}}}+C.}
O integralui
I
2
{\displaystyle I_{2}}
v
=
3
+
1
t
2
,
v
2
=
3
+
1
t
2
,
v
2
−
3
=
1
t
2
,
{\displaystyle v={\sqrt {3+{\frac {1}{t^{2}}}}},\;v^{2}=3+{\frac {1}{t^{2}}},\;\;v^{2}-3={\frac {1}{t^{2}}},}
d
(
1
t
2
)
=
d
(
v
2
−
3
)
=
2
v
d
v
;
{\displaystyle d{\Big (}{\frac {1}{t^{2}}}{\Big )}=d(v^{2}-3)=2v\;dv;}
I
2
=
2
∫
d
t
(
t
2
+
3
)
3
t
2
+
1
=
2
∫
d
t
t
2
(
1
+
3
1
t
2
)
t
3
+
1
t
2
=
2
∫
d
t
t
3
(
1
+
3
1
t
2
)
3
+
1
t
2
=
−
2
2
∫
d
(
1
t
2
)
(
1
+
3
1
t
2
)
3
+
1
t
2
=
{\displaystyle I_{2}=2\int {\frac {dt}{(t^{2}+3){\sqrt {3t^{2}+1}}}}=2\int {\frac {dt}{t^{2}(1+3{\frac {1}{t^{2}}})t{\sqrt {3+{\frac {1}{t^{2}}}}}}}=2\int {\frac {\frac {dt}{t^{3}}}{(1+3{\frac {1}{t^{2}}}){\sqrt {3+{\frac {1}{t^{2}}}}}}}=-{2 \over 2}\int {\frac {d({\frac {1}{t^{2}}})}{(1+3{\frac {1}{t^{2}}}){\sqrt {3+{\frac {1}{t^{2}}}}}}}=}
=
−
∫
2
v
d
v
(
1
+
3
(
v
2
−
3
)
)
⋅
v
=
−
2
∫
v
d
v
(
1
+
3
v
2
−
9
)
⋅
v
=
−
2
∫
d
v
3
v
2
−
8
=
−
2
∫
d
v
3
(
v
2
−
8
3
)
.
{\displaystyle =-\int {\frac {2v\;dv}{(1+3(v^{2}-3))\cdot v}}=-2\int {\frac {v\;dv}{(1+3v^{2}-9)\cdot v}}=-2\int {\frac {dv}{3v^{2}-8}}=-2\int {\frac {dv}{3(v^{2}-{\frac {8}{3}})}}.}
Toliau pasinaudosime racionaliųjų funkcijų angliškos Vikipedijos integralų lentele iš čia: https://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions
Mums prireiks šito integralo:
∫
1
x
2
−
a
2
d
x
=
1
2
a
ln
|
x
−
a
x
+
a
|
+
C
=
{
−
1
a
artanh
x
a
+
C
=
1
2
a
ln
a
−
x
a
+
x
+
C
(for
|
x
|
<
|
a
|
)
−
1
a
arcoth
x
a
+
C
=
1
2
a
ln
x
−
a
x
+
a
+
C
(for
|
x
|
>
|
a
|
)
{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}\,dx={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C={\begin{cases}\displaystyle -{\frac {1}{a}}\,\operatorname {artanh} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C&{\text{(for }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle -{\frac {1}{a}}\,\operatorname {arcoth} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C&{\text{(for }}|x|>|a|{\mbox{)}}\end{cases}}}
Tada, kai
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{\displaystyle |v|<{\sqrt {\frac {8}{3}}},}
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{\displaystyle I_{2}=-{\frac {2}{3}}\int {\frac {dv}{v^{2}-{\frac {8}{3}}}}=-{\frac {2}{3}}\cdot {\frac {1}{2{\sqrt {\frac {8}{3}}}}}\ln {\frac {{\sqrt {\frac {8}{3}}}-v}{{\sqrt {\frac {8}{3}}}+v}}+C=-{\frac {1}{{\sqrt {3}}\cdot {\sqrt {8}}}}\ln {\frac {{\sqrt {\frac {8}{3}}}-v}{{\sqrt {\frac {8}{3}}}+v}}+C={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {8}{3}}}+v}{{\sqrt {\frac {8}{3}}}-v}}+C=}
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{\displaystyle ={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {8}{3}}}+{\sqrt {3+{\frac {1}{t^{2}}}}}}{{\sqrt {\frac {8}{3}}}-{\sqrt {3+{\frac {1}{t^{2}}}}}}}+C={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {8}{3}}}+{\sqrt {3+{\frac {1}{({\frac {x+1}{1-x}})^{2}}}}}}{{\sqrt {\frac {8}{3}}}-{\sqrt {3+{\frac {1}{({\frac {x+1}{1-x}})^{2}}}}}}}+C={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {8}{3}}}+{\sqrt {3+{\frac {(1-x)^{2}}{(x+1)^{2}}}}}}{{\sqrt {\frac {8}{3}}}-{\sqrt {3+{\frac {(1-x)^{2}}{(x+1)^{2}}}}}}}+C=}
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{\displaystyle ={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {8}{3}}}+{\sqrt {\frac {3(x^{2}+2x+1)+(1-2x+x^{2})}{(x+1)^{2}}}}}{{\sqrt {\frac {8}{3}}}-{\sqrt {\frac {3x^{2}+6x+3+(1-2x+x^{2})}{(x+1)^{2}}}}}}+C={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {8}{3}}}+{\sqrt {\frac {4x^{2}+4x+4}{(x+1)^{2}}}}}{{\sqrt {\frac {8}{3}}}-{\sqrt {\frac {4x^{2}+4x+4}{(x+1)^{2}}}}}}+C={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {8}{3}}}+2{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}}{{\sqrt {\frac {8}{3}}}-2{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}}}+C=}
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{\displaystyle ={\frac {1}{2{\sqrt {6}}}}\ln {\frac {2{\sqrt {\frac {2}{3}}}+2{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}}{2{\sqrt {\frac {2}{3}}}-2{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}}}+C={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {2}{3}}}+{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}}{{\sqrt {\frac {2}{3}}}-{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}}}+C.}
Tada, kai
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{\displaystyle |v|>{\sqrt {\frac {8}{3}}},}
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{\displaystyle I_{2}={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}+{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}-{\sqrt {\frac {2}{3}}}}}+C.}
Na, o [labai seno] vadovelio atsakyme, panašu, kad įsivėlus klaidelė, nes gaunamas toks (truputi kitoks) atsakymas:
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{\displaystyle I_{2}=-2\int {\frac {dv}{3v^{2}-8}}={\frac {1}{2{\sqrt {6}}}}\ln {\Big |}{\frac {v+{\sqrt {\frac {8}{3}}}}{v-{\sqrt {\frac {8}{3}}}}}{\Big |}+C={\frac {1}{2{\sqrt {6}}}}\ln {\Big |}{\frac {{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}+{\sqrt {\frac {8}{3}}}}{{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}-{\sqrt {\frac {8}{3}}}}}{\Big |}+C.}
Vadovelio atsakymas skiriasi tuom, kad turi modulio ženklą ir
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{\displaystyle {\sqrt {\frac {8}{3}}}}
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{\displaystyle {\sqrt {\frac {2}{3}}}.}
![{\displaystyle R{\Big (}x,\;{\sqrt[{n}]{\frac {ax+b}{cx+d}}}{\Big )}\quad (7.64)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11ef778b739252c080a6f21f13145790bdfc3c50)
![{\displaystyle t={\sqrt[{n}]{\frac {ax+b}{cx+d}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d18efee0a15dfe22373524d345f001ef8ef0cf9d)
![{\displaystyle \int R{\Big (}x,\;{\sqrt[{n}]{\frac {ax+b}{cx+d}}}{\Big )}\;dx=\int R{\Big (}{\frac {dt^{n}-b}{a-ct^{n}}},\;t{\Big )}\;{\frac {(ad-bc)nt^{n-1}}{(a-ct^{n})^{2}}}\;dt.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9607991f23093b7bda4f9dd9b44948254504fe2)
![{\displaystyle \int {\frac {(Mx+N)\;dx}{(x^{2}+px+q)^{\lambda }{\sqrt {ax^{2}+bx+c}}}}=\int {\frac {Mt+{\Big (}{\frac {-Mp}{2}}+N{\Big )}}{{\Big [}t^{2}+{\Big (}q-{\frac {p^{2}}{4}}{\Big )}{\Big ]}^{\lambda }{\sqrt {at^{2}+{\Big (}c-{\frac {ap^{2}}{4}}{\Big )}}}}}\;dt.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/003290f0521c0dc7b89b62612ff5c6b4ccb119eb)
![{\displaystyle \int {\frac {Mt+{\Big (}{\frac {-Mp}{2}}+N{\Big )}}{{\Big [}t^{2}+{\Big (}q-{\frac {p^{2}}{4}}{\Big )}{\Big ]}^{\lambda }{\sqrt {at^{2}+{\Big (}c-{\frac {ap^{2}}{4}}{\Big )}}}}}\;dt=\int {\frac {M(x+{\frac {p}{2}})+{\Big (}{\frac {-Mp}{2}}+N{\Big )}}{{\Big [}(x+{\frac {p}{2}})^{2}+{\Big (}q-{\frac {p^{2}}{4}}{\Big )}{\Big ]}^{\lambda }{\sqrt {a(x+{\frac {p}{2}})^{2}+{\Big (}c-{\frac {ap^{2}}{4}}{\Big )}}}}}\;dx=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a73aed8f0afcf548845a14c8ee8aa376a3bd196)
![{\displaystyle =\int {\frac {Mx+{\frac {Mp}{2}}+{\frac {-Mp}{2}}+N}{{\Big [}x^{2}+px+{\frac {p^{2}}{4}}+{\Big (}q-{\frac {p^{2}}{4}}{\Big )}{\Big ]}^{\lambda }{\sqrt {a(x^{2}+px+{\frac {p^{2}}{4}})+{\Big (}c-{\frac {ap^{2}}{4}}{\Big )}}}}}\;dx=\int {\frac {Mx+N}{{\Big [}x^{2}+px+q{\Big ]}^{\lambda }{\sqrt {ax^{2}+apx+{\frac {ap^{2}}{4}}+c-{\frac {ap^{2}}{4}}}}}}\;dx=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80894f350056825b79683e2cecd98bc226c3b2ec)
![{\displaystyle =\int {\frac {Mx+N}{[x^{2}+px+q]^{\lambda }{\sqrt {ax^{2}+apx+c}}}}\;dx=\int {\frac {(Mx+N)\;dx}{(x^{2}+px+q)^{\lambda }{\sqrt {ax^{2}+bx+c}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aff1383b8f51263c4fedb2c6ac2ad1ce2d64f25c)
![{\displaystyle x^{2}+px+q={\frac {(\mu ^{2}+p\mu +q)t^{2}+[2\mu \nu +p(\mu +\nu )+2q]t+(\nu ^{2}+p\nu +q)}{(1+t)^{2}}}\;,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cce803cbf64e894c2107bf31e24ea85f718acdc7)
![{\displaystyle ax^{2}+bx+c={\frac {(a\mu ^{2}+b\mu +c)t^{2}+[2\mu \nu a+b(\mu +\nu )+2c]t+(a\nu ^{2}+b\nu +c)}{(1+t)^{2}}}\;.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfa8d2689b8f867218340ec9cd9bbaee52747fb6)
![{\displaystyle [2(c+aq)-bp]^{2}>(4q-p^{2})(4ac-b^{2}).\quad (7.79)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e77d804df57ccabb774c9fcf1c82b20d5e763431)
![{\displaystyle [2(c+aq)-bp]^{2}\geq [4{\sqrt {qac}}-bp]^{2}=(4q-p^{2})(4ac-b^{2})+4(p{\sqrt {ac}}-b{\sqrt {q}})^{2}\geq (4q-p^{2})(4ac-b^{2}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c433a1d8813014e3c0f4f6ebfefcc5ef02165953)
![{\displaystyle x^{2}+x+1={\frac {(\mu ^{2}+\mu +1)t^{2}+[2\mu \nu +(\mu +\nu )+2]t+(\nu ^{2}+\nu +1)}{(1+t)^{2}}}\;,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b0ddd873d3b5123eb46819b9631cd90c2b37cb9)
![{\displaystyle x^{2}-x+1={\frac {(\mu ^{2}-\mu +1)t^{2}+[2\mu \nu -(\mu +\nu )+2]t+(\nu ^{2}-\nu +1)}{(1+t)^{2}}}\;.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d57cec0a8425d1944e1e94da34153d14055539d2)
![{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}\,dx={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C={\begin{cases}\displaystyle -{\frac {1}{a}}\,\operatorname {artanh} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C&{\text{(for }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle -{\frac {1}{a}}\,\operatorname {arcoth} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C&{\text{(for }}|x|>|a|{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f707d02dbc04ceb2d02ddb5bfd60ab31f45b6b55)
