Tarkime, kad funkcijos
u
(
x
)
{\displaystyle u(x)}
v
(
x
)
{\displaystyle v(x)}
išvestines . Tada:
∫
u
(
x
)
v
′
(
x
)
d
x
=
u
(
x
)
v
(
x
)
−
∫
v
(
x
)
u
′
(
x
)
d
x
.
{\displaystyle \int u(x)v'(x){\mathsf {d}}x=u(x)v(x)-\int v(x)u'(x){\mathsf {d}}x.}
Lygtis nesunkiai įrodoma prisiminus sandaugos diferenciavimo taisyklę:
∫
d
(
u
v
)
=
u
v
,
{\displaystyle \int {\mathsf {d}}(uv)=uv,}
∫
u
d
v
+
∫
v
d
u
=
u
v
,
{\displaystyle \int u{\mathsf {d}}v+\int v{\mathsf {d}}u=uv,}
∫
u
d
v
=
u
v
−
∫
v
d
u
.
{\displaystyle \int u{\mathsf {d}}v=uv-\int v{\mathsf {d}}u.}
∫
x
e
x
d
x
=
x
e
x
−
∫
e
x
d
x
=
x
e
x
−
e
x
+
C
.
{\displaystyle \int x{\mathsf {e}}^{x}{\mathsf {d}}x=x{\mathsf {e}}^{x}-\int {\mathsf {e}}^{x}{\mathsf {d}}x=x{\mathsf {e}}^{x}-{\mathsf {e}}^{x}+C.}
Čia
u
(
x
)
=
x
{\displaystyle u(x)=x}
v
′
(
x
)
=
e
x
{\displaystyle v'(x)=e^{x}}
∫
x
n
ln
x
d
x
{\displaystyle \int x^{n}\ln x{\mathsf {d}}x}
u = ln x, u'=(ln x)'=1/x,
v
=
∫
x
n
d
x
=
x
n
+
1
n
+
1
{\displaystyle v=\int x^{n}{\mathsf {d}}x={\frac {x^{n+1}}{n+1}}}
v
′
=
(
x
n
+
1
n
+
1
)
′
=
x
n
{\displaystyle v'=({\frac {x^{n+1}}{n+1}})'=x^{n}}
∫
ln
(
x
)
x
n
d
x
=
ln
x
x
n
+
1
n
+
1
−
∫
x
n
+
1
n
+
1
1
x
d
x
=
ln
x
x
n
+
1
n
+
1
−
1
n
+
1
∫
x
n
d
x
=
x
n
+
1
n
+
1
(
ln
x
−
1
n
+
1
)
+
C
.
{\displaystyle \int \ln(x)x^{n}{\mathsf {d}}x=\ln x{\frac {x^{n+1}}{n+1}}-\int {\frac {x^{n+1}}{n+1}}{\frac {1}{x}}{\mathsf {d}}x=\ln x{\frac {x^{n+1}}{n+1}}-{\frac {1}{n+1}}\int x^{n}{\mathsf {d}}x={\frac {x^{n+1}}{n+1}}(\ln x-{\frac {1}{n+1}})+C.}
∫
x
sin
x
d
x
{\displaystyle \int x\sin xdx}
u=x, dv=sin(x) dx, du=dx,
v
=
∫
sin
x
d
x
=
−
cos
x
.
{\displaystyle v=\int \sin xdx=-\cos x.}
∫
u
d
v
=
u
v
−
∫
v
d
u
=
−
x
cos
x
−
∫
−
cos
x
d
x
=
−
x
cos
x
+
sin
x
+
C
.
{\displaystyle \int udv=uv-\int vdu=-x\cos x-\int -\cos xdx=-x\cos x+\sin x+C.}
∫
arcsin
x
d
x
.
{\displaystyle \int \arcsin xdx.}
u=arcsin(x); dv=dx;
d
u
=
d
x
1
−
x
2
;
{\displaystyle du={\frac {dx}{\sqrt {1-x^{2}}}};}
x
arcsin
x
−
∫
x
d
x
1
−
x
2
=
x
arcsin
x
+
∫
d
(
1
−
x
2
)
2
1
−
x
2
=
x
arcsin
x
+
1
−
x
2
+
C
.
{\displaystyle x\arcsin x-\int {\frac {xdx}{\sqrt {1-x^{2}}}}=x\arcsin x+\int {\frac {d(1-x^{2})}{2{\sqrt {1-x^{2}}}}}=x\arcsin x+{\sqrt {1-x^{2}}}+C.}
Patikriname
(
x
arcsin
x
+
1
−
x
2
+
C
)
′
=
arcsin
(
x
)
+
x
/
(
1
−
x
2
)
0.5
+
0.5
⋅
(
−
2
x
)
/
(
1
−
x
2
)
0.5
=
arcsin
x
.
{\displaystyle (x\arcsin x+{\sqrt {1-x^{2}}}+C)'=\arcsin(x)+x/(1-x^{2})^{0.5}+0.5\cdot (-2x)/(1-x^{2})^{0.5}=\arcsin x.}
∫
x
ln
x
d
x
=
x
2
2
ln
x
−
∫
x
2
2
1
x
d
x
=
x
2
2
ln
x
−
x
2
4
+
C
,
{\displaystyle \int x\ln xdx={\frac {x^{2}}{2}}\ln x-\int {\frac {x^{2}}{2}}{\frac {1}{x}}dx={\frac {x^{2}}{2}}\ln x-{\frac {x^{2}}{4}}+C,}
kur u=ln(x); dv=x dx; du=1/x dx;
v
=
∫
x
d
x
=
x
2
2
.
{\displaystyle v=\int xdx={\frac {x^{2}}{2}}.}
∫
ln
x
d
x
=
x
ln
x
−
∫
x
d
x
x
=
x
ln
x
−
x
+
C
,
{\displaystyle \int \ln xdx=x\ln x-\int x{\frac {dx}{x}}=x\ln x-x+C,}
kur u=ln(x); dv=dx; du=1/x dx; v=x.
∫
arctan
x
d
x
=
x
arctan
x
−
∫
x
d
x
1
+
x
2
=
x
arctan
x
−
1
2
∫
d
(
1
+
x
2
)
1
+
x
2
=
x
arctan
x
−
1
2
ln
(
1
+
x
2
)
+
C
{\displaystyle \int \arctan xdx=x\arctan x-\int {\frac {xdx}{1+x^{2}}}=x\arctan x-{\frac {1}{2}}\int {\frac {d(1+x^{2})}{1+x^{2}}}=x\arctan x-{\frac {1}{2}}\ln(1+x^{2})+C}
kur u=arctg(x); dv=dx;
d
u
=
1
1
+
x
2
d
x
;
{\displaystyle du={\frac {1}{1+x^{2}}}dx;}
∫
x
2
e
x
d
x
=
x
2
e
x
−
2
∫
x
e
x
d
x
=
x
2
e
x
−
2
(
x
−
1
)
e
x
+
2
C
1
=
e
x
(
x
2
−
2
x
+
2
)
+
2
C
1
,
{\displaystyle \int x^{2}e^{x}dx=x^{2}e^{x}-2\int xe^{x}dx=x^{2}e^{x}-2(x-1)e^{x}+2C_{1}=e^{x}(x^{2}-2x+2)+2C_{1},}
kur
u
=
x
2
;
{\displaystyle u=x^{2};}
d
v
=
e
x
d
x
{\displaystyle dv=e^{x}dx}
v
=
e
x
{\displaystyle v=e^{x}}
∫
x
e
x
d
x
=
x
e
x
−
∫
x
′
e
x
d
x
=
x
e
x
−
e
x
+
C
1
.
{\displaystyle \int xe^{x}dx=xe^{x}-\int x'e^{x}dx=xe^{x}-e^{x}+C_{1}.}
I
=
∫
e
x
sin
x
d
x
.
{\displaystyle I=\int e^{x}\sin xdx.}
u
=
e
x
,
{\displaystyle u=e^{x},}
d
u
=
e
x
d
x
,
{\displaystyle du=e^{x}dx,}
v
=
∫
sin
x
d
x
=
−
cos
x
;
{\displaystyle v=\int \sin xdx=-\cos x;}
I
=
−
e
x
cos
x
+
∫
e
x
cos
x
d
x
.
{\displaystyle I=-e^{x}\cos x+\int e^{x}\cos xdx.}
Dar kartą integruojame dalimis.
u
=
e
x
,
{\displaystyle u=e^{x},}
d
u
=
e
x
d
x
,
{\displaystyle du=e^{x}dx,}
v
=
∫
cos
x
d
x
=
sin
x
;
{\displaystyle v=\int \cos xdx=\sin x;}
I
=
−
e
x
cos
x
+
e
x
sin
x
−
∫
e
x
sin
x
d
x
.
{\displaystyle I=-e^{x}\cos x+e^{x}\sin x-\int e^{x}\sin xdx.}
Turime, kad
I
=
e
x
(
sin
x
−
cos
x
)
−
I
.
{\displaystyle I=e^{x}(\sin x-\cos x)-I.}
2
I
=
e
x
(
sin
x
−
cos
x
)
{\displaystyle 2I=e^{x}(\sin x-\cos x)}
I
=
1
2
e
x
(
sin
x
−
cos
x
)
+
C
.
{\displaystyle I={\frac {1}{2}}e^{x}(\sin x-\cos x)+C.}
∫
8
x
3
ln
x
d
x
=
8
(
x
4
4
ln
x
−
∫
x
4
4
⋅
1
x
d
x
)
=
x
4
(
2
ln
x
−
1
2
)
+
C
,
{\displaystyle \int 8x^{3}\ln xdx=8({\frac {x^{4}}{4}}\ln x-\int {\frac {x^{4}}{4}}\cdot {\frac {1}{x}}dx)=x^{4}(2\ln x-{\frac {1}{2}})+C,}
u
=
ln
x
;
{\displaystyle u=\ln x;}
v
′
=
x
3
;
{\displaystyle v'=x^{3};}
u
′
=
1
x
;
{\displaystyle u'={\frac {1}{x}};}
v
=
x
4
4
.
{\displaystyle v={\frac {x^{4}}{4}}.}
∫
x
5
cos
x
3
d
x
,
{\displaystyle \int x^{5}\cos x^{3}dx,}
d
t
=
d
(
x
3
)
=
3
x
2
d
x
;
{\displaystyle dt=d(x^{3})=3x^{2}dx;}
d
x
=
d
t
3
x
2
;
{\displaystyle dx={\frac {dt}{3x^{2}}};}
∫
x
5
cos
x
3
d
x
=
∫
x
5
cos
x
3
d
(
x
3
)
3
x
2
=
1
3
∫
x
3
cos
x
3
d
(
x
3
)
=
1
3
∫
t
cos
t
d
t
=
1
3
(
t
sin
t
−
∫
sin
t
d
t
)
=
{\displaystyle \int x^{5}\cos x^{3}dx=\int x^{5}\cos x^{3}{\frac {d(x^{3})}{3x^{2}}}={\frac {1}{3}}\int x^{3}\cos x^{3}d(x^{3})={\frac {1}{3}}\int t\cos tdt={\frac {1}{3}}(t\sin t-\int \sin tdt)=}
=
1
3
(
t
sin
t
+
cos
t
)
+
C
=
1
3
(
x
3
sin
x
3
+
cos
x
3
)
+
C
,
{\displaystyle ={\frac {1}{3}}(t\sin t+\cos t)+C={\frac {1}{3}}(x^{3}\sin x^{3}+\cos x^{3})+C,}
v
′
=
cos
t
;
{\displaystyle v'=\cos t;}
u
=
t
;
{\displaystyle u=t;}
v
=
sin
t
;
{\displaystyle v=\sin t;}
u
′
=
1.
{\displaystyle u'=1.}
∫
arctan
x
d
x
=
x
arctan
x
−
∫
x
d
x
1
+
x
2
=
x
arctan
x
−
1
2
∫
d
(
1
+
x
2
)
1
+
x
2
=
{\displaystyle \int \arctan x\;dx=x\arctan x-\int {x\;dx \over 1+x^{2}}=x\arctan x-{1 \over 2}\int {d(1+x^{2}) \over 1+x^{2}}=}
=
x
arctan
x
−
1
2
ln
|
1
+
x
2
|
+
C
,
{\displaystyle =x\arctan x-{1 \over 2}\ln |1+x^{2}|+C,}
u
=
arctan
x
;
{\displaystyle u=\arctan x;}
d
v
=
d
x
;
{\displaystyle dv=dx;}
d
u
=
d
x
1
+
x
2
;
{\displaystyle du={dx \over 1+x^{2}};}
v
=
x
;
{\displaystyle v=x;}
d
(
1
+
x
2
)
=
2
x
d
x
.
{\displaystyle d(1+x^{2})=2xdx.}
Apskaičiuosime integralą
I
=
∫
x
arctan
x
d
x
.
{\displaystyle I=\int x\arctan x\;dx.}
u
=
arctan
x
,
d
v
=
x
d
x
,
{\displaystyle u=\arctan x,\;\;dv=x\;dx,}
d
u
=
d
x
1
+
x
2
,
v
=
x
2
2
.
{\displaystyle du={\frac {dx}{1+x^{2}}},\;\;v={\frac {x^{2}}{2}}.}
∫
u
d
v
=
u
v
−
∫
v
d
u
{\displaystyle \int u{\mathsf {d}}v=uv-\int v{\mathsf {d}}u}
I
=
x
2
2
arctan
x
−
1
2
∫
x
2
1
+
x
2
d
x
=
x
2
2
arctan
x
−
1
2
∫
(
1
+
x
2
)
−
1
1
+
x
2
d
x
=
{\displaystyle I={\frac {x^{2}}{2}}\arctan x-{\frac {1}{2}}\int {\frac {x^{2}}{1+x^{2}}}dx={\frac {x^{2}}{2}}\arctan x-{\frac {1}{2}}\int {\frac {(1+x^{2})-1}{1+x^{2}}}dx=}
=
x
2
2
arctan
x
−
1
2
∫
d
x
+
1
2
∫
d
x
1
+
x
2
=
x
2
2
arctan
x
−
x
2
+
arctan
x
2
+
C
=
x
2
+
1
2
arctan
x
−
x
2
+
C
.
{\displaystyle ={\frac {x^{2}}{2}}\arctan x-{\frac {1}{2}}\int dx+{\frac {1}{2}}\int {\frac {dx}{1+x^{2}}}={\frac {x^{2}}{2}}\arctan x-{\frac {x}{2}}+{\frac {\arctan x}{2}}+C={\frac {x^{2}+1}{2}}\arctan x-{\frac {x}{2}}+C.}
Apskaičiuosime integralą
I
=
∫
x
2
cos
x
d
x
.
{\displaystyle I=\int x^{2}\cos x\;dx.}
u
=
x
2
,
d
v
=
cos
x
d
x
,
{\displaystyle u=x^{2},\;\;dv=\cos x\;dx,}
d
u
=
2
x
d
x
,
v
=
sin
x
.
{\displaystyle du=2x\;dx,\;\;v=\sin x.}
I
=
x
2
sin
x
−
2
∫
x
sin
x
d
x
.
{\displaystyle I=x^{2}\sin x-2\int x\sin x\;dx.}
Paskutinį integralą vėl skaičiuosime pagal
∫
u
d
v
=
u
v
−
∫
v
d
u
{\displaystyle \int u{\mathsf {d}}v=uv-\int v{\mathsf {d}}u}
u
=
x
,
d
v
=
sin
x
d
x
,
{\displaystyle u=x,\;\;dv=\sin x\;dx,}
d
u
=
d
x
,
v
=
−
cos
x
;
{\displaystyle du=dx,\;\;v=-\cos x;}
I
=
x
2
sin
x
−
2
(
−
x
cos
x
−
∫
−
cos
x
d
x
)
=
{\displaystyle I=x^{2}\sin x-2(-x\cos x-\int -\cos x\;dx)=}
=
x
2
sin
x
+
2
x
cos
x
−
2
∫
cos
x
d
x
=
(
x
2
−
2
)
sin
x
+
2
x
cos
x
+
C
.
{\displaystyle =x^{2}\sin x+2x\cos x-2\int \cos x\;dx=(x^{2}-2)\sin x+2x\cos x+C.}
Vadinasi, integralą
∫
x
2
cos
x
d
x
{\displaystyle \int x^{2}\cos x\;dx}
∫
x
n
cos
x
d
x
{\displaystyle \int x^{n}\cos x\;dx}
n - natūrinis skaičius) galima apskaičiuoti analogiškai, taikant dalinio integravimo formulę n kartų.
Dabar apskaičiuosime integralą
I
=
∫
e
a
x
cos
(
b
x
)
d
x
{\displaystyle I=\int e^{ax}\cos(bx)dx}
a =const, b =const). Iš pradžių pritaikysime formulę, tarę, kad
u
=
e
a
x
,
d
v
=
cos
b
x
d
x
.
{\displaystyle u=e^{ax},\;\;dv=\cos bx\;dx.}
d
u
=
a
e
a
x
d
x
,
v
=
sin
b
x
b
,
{\displaystyle du=ae^{ax}dx,\;\;v={\frac {\sin bx}{b}},}
I
=
e
a
x
sin
b
x
b
−
a
b
∫
e
a
x
sin
(
b
x
)
d
x
.
{\displaystyle I={\frac {e^{ax}\sin bx}{b}}-{\frac {a}{b}}\int e^{ax}\sin(bx)dx.}
Paskutinį integralą skaičiuojame vėl pagal integravimo dalimis formulę, tik šį kartą
u
=
e
a
x
,
d
v
=
sin
b
x
d
x
.
{\displaystyle u=e^{ax},\;\;dv=\sin bx\;dx.}
d
u
=
a
e
a
x
d
x
,
v
=
−
cos
b
x
b
,
{\displaystyle du=ae^{ax}dx,\;\;v=-{\frac {\cos bx}{b}},}
I
=
e
a
x
sin
b
x
b
+
a
b
2
e
a
x
cos
(
b
x
)
−
a
2
b
2
∫
e
a
x
cos
(
b
x
)
d
x
,
{\displaystyle I={\frac {e^{ax}\sin bx}{b}}+{\frac {a}{b^{2}}}e^{ax}\cos(bx)-{\frac {a^{2}}{b^{2}}}\int e^{ax}\cos(bx)dx,}
I
=
e
a
x
sin
b
x
b
+
a
b
2
e
a
x
cos
(
b
x
)
−
a
2
b
2
I
,
(
6.11
)
{\displaystyle I={\frac {e^{ax}\sin bx}{b}}+{\frac {a}{b^{2}}}e^{ax}\cos(bx)-{\frac {a^{2}}{b^{2}}}I,\quad (6.11)}
I
+
a
2
b
2
I
=
e
a
x
sin
b
x
b
+
a
b
2
e
a
x
cos
(
b
x
)
,
{\displaystyle I+{\frac {a^{2}}{b^{2}}}I={\frac {e^{ax}\sin bx}{b}}+{\frac {a}{b^{2}}}e^{ax}\cos(bx),}
I
b
2
+
I
a
2
b
2
=
b
e
a
x
sin
b
x
+
a
e
a
x
cos
(
b
x
)
b
2
,
{\displaystyle {\frac {Ib^{2}+Ia^{2}}{b^{2}}}={\frac {be^{ax}\sin bx+ae^{ax}\cos(bx)}{b^{2}}},}
b
2
+
a
2
b
2
I
=
b
sin
b
x
+
a
cos
b
x
b
2
e
a
x
,
{\displaystyle {\frac {b^{2}+a^{2}}{b^{2}}}I={\frac {b\sin bx+a\cos bx}{b^{2}}}e^{ax},}
I
=
a
cos
b
x
+
b
sin
b
x
a
2
+
b
2
e
a
x
.
{\displaystyle I={\frac {a\cos bx+b\sin bx}{a^{2}+b^{2}}}e^{ax}.}
Vadinasi, du kartus pritaikę dalinio integravimo formulę, gavome pirmojo laipsnio (6.11) lygtį integralo I atžvilgiu. Iš tos lygties radome I .
Apskaičiuosime integralą
I
=
x
d
x
cos
2
x
.
{\displaystyle I={\frac {x\;dx}{\cos ^{2}x}}.}
∫
u
d
v
=
u
v
−
∫
v
d
u
{\displaystyle \int u\;{\mathsf {d}}v=uv-\int v\;{\mathsf {d}}u}
u
=
x
,
d
v
=
d
x
cos
2
x
.
{\displaystyle u=x,\;\;dv={\frac {dx}{\cos ^{2}x}}.}
du=dx , v=tg x ,
I
=
x
tan
x
−
∫
tan
x
d
x
=
x
tan
x
−
∫
sin
x
d
x
cos
x
=
x
tan
x
+
∫
d
(
cos
x
)
cos
x
=
x
tan
x
+
ln
|
cos
x
|
+
C
.
{\displaystyle I=x\tan x-\int \tan x\;dx=x\tan x-\int {\frac {\sin x\;dx}{\cos x}}=x\tan x+\int {\frac {d(\cos x)}{\cos x}}=x\tan x+\ln |\cos x|+C.}
I
=
∫
x
2
+
a
2
d
x
.
{\displaystyle I=\int {\sqrt {x^{2}+a^{2}}}\;dx.}
u
=
x
2
+
a
2
,
d
u
=
x
x
2
+
a
2
d
x
,
d
v
=
d
x
,
v
=
∫
d
x
=
x
;
{\displaystyle u={\sqrt {x^{2}+a^{2}}},\;\;du={\frac {x}{\sqrt {x^{2}+a^{2}}}}dx,\;\;dv=dx,\;\;v=\int dx=x;}
I
=
∫
x
2
+
a
2
d
x
=
u
v
−
∫
v
d
u
=
x
x
2
+
a
2
−
∫
x
2
x
2
+
a
2
d
x
=
x
x
2
+
a
2
−
∫
(
x
2
+
a
2
)
−
a
2
x
2
+
a
2
d
x
=
{\displaystyle I=\int {\sqrt {x^{2}+a^{2}}}\;dx=uv-\int v\;du=x{\sqrt {x^{2}+a^{2}}}-\int {\frac {x^{2}}{\sqrt {x^{2}+a^{2}}}}dx=x{\sqrt {x^{2}+a^{2}}}-\int {\frac {(x^{2}+a^{2})-a^{2}}{\sqrt {x^{2}+a^{2}}}}dx=}
=
x
x
2
+
a
2
−
∫
x
2
+
a
2
d
x
+
a
2
∫
d
x
x
2
+
a
2
=
x
x
2
+
a
2
−
∫
x
2
+
a
2
d
x
+
a
2
ln
(
x
+
x
2
+
a
2
)
.
{\displaystyle =x{\sqrt {x^{2}+a^{2}}}-\int {\sqrt {x^{2}+a^{2}}}\;dx+a^{2}\int {\frac {dx}{\sqrt {x^{2}+a^{2}}}}=x{\sqrt {x^{2}+a^{2}}}-\int {\sqrt {x^{2}+a^{2}}}\;dx+a^{2}\ln(x+{\sqrt {x^{2}+a^{2}}}).}
I
=
x
x
2
+
a
2
−
I
+
a
2
ln
(
x
+
x
2
+
a
2
)
.
{\displaystyle I=x{\sqrt {x^{2}+a^{2}}}-I+a^{2}\ln(x+{\sqrt {x^{2}+a^{2}}}).}
Taigi dešinėje pusėje gavome pradinį integralą, kurį perkėlę į kairiąją pusę, turėsime:
2
I
=
x
x
2
+
a
2
+
a
2
ln
(
x
+
x
2
+
a
2
)
.
{\displaystyle 2I=x{\sqrt {x^{2}+a^{2}}}+a^{2}\ln(x+{\sqrt {x^{2}+a^{2}}}).}
Todėl
∫
x
2
+
a
2
d
x
=
x
2
x
2
+
a
2
+
a
2
2
ln
(
x
+
x
2
+
a
2
)
+
C
.
{\displaystyle \int {\sqrt {x^{2}+a^{2}}}\;dx={\frac {x}{2}}{\sqrt {x^{2}+a^{2}}}+{\frac {a^{2}}{2}}\ln(x+{\sqrt {x^{2}+a^{2}}})+C.}
Analogiškai gautume:
∫
x
2
−
a
2
d
x
=
x
2
x
2
−
a
2
−
a
2
2
ln
(
x
+
x
2
−
a
2
)
+
C
.
{\displaystyle \int {\sqrt {x^{2}-a^{2}}}\;dx={\frac {x}{2}}{\sqrt {x^{2}-a^{2}}}-{\frac {a^{2}}{2}}\ln(x+{\sqrt {x^{2}-a^{2}}})+C.}
∫
x
ln
x
d
x
.
{\displaystyle \int {\sqrt {x}}\ln x\;dx.}
t
=
x
.
{\displaystyle t={\sqrt {x}}.}
t
2
=
x
,
2
t
d
t
=
d
x
.
{\displaystyle t^{2}=x,\;\;2t\;dt=dx.}
∫
x
ln
x
d
x
=
∫
t
ln
(
t
2
)
⋅
2
t
d
t
=
2
∫
t
2
ln
(
t
2
)
d
t
=
4
∫
t
2
ln
t
d
t
.
{\displaystyle \int {\sqrt {x}}\ln x\;dx=\int t\ln(t^{2})\cdot 2t\;dt=2\int t^{2}\ln(t^{2})\;dt=4\int t^{2}\ln t\;dt.}
Tokį integralą integruoti dalimis jau mokame. Pažymime
u
=
ln
t
,
d
v
=
t
2
d
t
,
d
u
=
d
t
t
,
v
=
∫
t
2
d
t
=
t
3
3
.
{\displaystyle u=\ln t,\;\;dv=t^{2}dt,\;\;du={dt \over t},\;\;v=\int t^{2}dt={\frac {t^{3}}{3}}.}
4
∫
t
2
ln
t
d
t
=
4
(
u
v
−
∫
v
d
u
)
=
4
(
t
3
3
ln
t
−
∫
t
3
3
d
t
t
)
=
4
(
t
3
3
ln
t
−
1
3
∫
t
2
d
t
)
=
{\displaystyle 4\int t^{2}\ln t\;dt=4(uv-\int v\;du)=4{\Big (}{\frac {t^{3}}{3}}\ln t-\int {\frac {t^{3}}{3}}{dt \over t}{\Big )}=4{\Big (}{\frac {t^{3}}{3}}\ln t-{\frac {1}{3}}\int t^{2}\;dt{\Big )}=}
=
4
(
t
3
3
ln
t
−
t
3
9
)
+
C
=
4
x
3
2
3
ln
x
−
4
x
3
2
9
+
C
=
2
x
3
2
3
ln
x
−
4
x
3
2
9
+
C
.
{\displaystyle =4{\Big (}{\frac {t^{3}}{3}}\ln t-{\frac {t^{3}}{9}}{\Big )}+C={\frac {4x^{3 \over 2}}{3}}\ln {\sqrt {x}}-{\frac {4x^{3 \over 2}}{9}}+C={\frac {2x^{3 \over 2}}{3}}\ln x-{\frac {4x^{3 \over 2}}{9}}+C.}
∫
x
ln
x
d
x
.
{\displaystyle \int {\sqrt {x}}\ln x\;dx.}
Pažymime
u
=
ln
x
,
d
v
=
x
d
x
,
d
u
=
d
x
x
,
v
=
∫
x
1
/
2
d
x
=
x
3
/
2
3
/
2
=
2
3
x
3
/
2
.
{\displaystyle u=\ln x,\;\;dv={\sqrt {x}}\;dx,\;\;du={dx \over x},\;\;v=\int x^{1/2}dx={\frac {x^{3/2}}{3/2}}={\frac {2}{3}}x^{3/2}.}
∫
x
ln
x
d
x
=
(
u
v
−
∫
v
d
u
)
=
2
3
x
3
/
2
ln
x
−
∫
2
3
x
3
/
2
d
x
x
=
2
3
x
3
/
2
ln
x
−
2
3
∫
x
1
/
2
d
x
=
{\displaystyle \int {\sqrt {x}}\ln x\;dx=(uv-\int v\;du)={\frac {2}{3}}x^{3/2}\ln x-\int {\frac {2}{3}}x^{3/2}{dx \over x}={\frac {2}{3}}x^{3/2}\ln x-{\frac {2}{3}}\int x^{1/2}dx=}
=
2
3
x
3
/
2
ln
x
−
2
3
x
3
/
2
3
/
2
+
C
=
2
3
x
3
/
2
ln
x
−
4
9
x
3
/
2
+
C
.
{\displaystyle ={\frac {2}{3}}x^{3/2}\ln x-{\frac {2}{3}}{\frac {x^{3/2}}{3/2}}+C={\frac {2}{3}}x^{3/2}\ln x-{\frac {4}{9}}x^{3/2}+C.}
![{\displaystyle K_{\lambda }={\frac {1}{a^{2}}}\int {\frac {a^{2}dt}{(t^{2}+a^{2})^{\lambda }}}={\frac {1}{a^{2}}}\int {\frac {[(t^{2}+a^{2})-t^{2}]dt}{(t^{2}+a^{2})^{\lambda }}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30b3e6d4178d7810756c728c44e9f313eb6b455a)
![{\displaystyle S=\int _{0}^{6}{\frac {dt}{(t^{2}+a^{2})^{4}}}={\Big [}{\frac {t}{6a^{2}(t^{2}+a^{2})^{3}}}+{\frac {5t}{24a^{4}(t^{2}+a^{2})^{2}}}+{\frac {5t}{16a^{6}(t^{2}+a^{2})}}+{\frac {5}{16a^{7}}}\arctan {\frac {t}{a}}{\Big ]}_{0}^{6}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00e664585b75abd84f0ec55d3a5da2076587e153)
![{\displaystyle ={\Big [}{\frac {6}{6\cdot 8^{2}(6^{2}+8^{2})^{3}}}+{\frac {5\cdot 6}{24\cdot 8^{4}(6^{2}+8^{2})^{2}}}+{\frac {5\cdot 6}{16\cdot 8^{6}(6^{2}+8^{2})}}+{\frac {5}{16\cdot 8^{7}}}\arctan {\frac {6}{8}}{\Big ]}-[0+{\frac {5}{16\cdot 8^{7}}}\arctan 0]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e27c5d8c987583e514f0f44b6675dd5accbfc22)
![{\displaystyle ={\Big [}{\frac {1}{64(36+64)^{3}}}+{\frac {5}{4\cdot 4096(36+64)^{2}}}+{\frac {5\cdot 3}{8\cdot 262144(36+64)}}+{\frac {5}{16\cdot 2097152}}\arctan {\frac {3}{4}}{\Big ]}-0=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c34f87389dc7bce3c487f7047d9bdd73602d4857)
