Tarkime, kad funkcijos u ( x ) {\displaystyle u(x)} ir v ( x ) {\displaystyle v(x)} turi tolydžias 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} , o 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}}}};} v=x.
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;} v=x.
∫ 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} ; du=2x 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},} dv=sin(x)dx; 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},} dv=cos(x)dx; 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.} Vadinasi 2 I = e x ( sin x − cos x ) {\displaystyle 2I=e^{x}(\sin x-\cos x)} , tai
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,} kur 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,} kur 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,} kur 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,}
kur 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.} Jei u = arctan x , d v = x d x , {\displaystyle u=\arctan x,\;\;dv=x\;dx,} tai d u = d x 1 + x 2 , v = x 2 2 . {\displaystyle du={\frac {dx}{1+x^{2}}},\;\;v={\frac {x^{2}}{2}}.} Todėl pagal formulę (∫ 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.} Jei u = x 2 , d v = cos x d x , {\displaystyle u=x^{2},\;\;dv=\cos x\;dx,} tai d u = 2 x d x , v = sin x . {\displaystyle du=2x\;dx,\;\;v=\sin x.} Tada, remiantis formule, 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} formulę. Jei šį kartą u = x , d v = sin x d x , {\displaystyle u=x,\;\;dv=\sin x\;dx,} tai d u = d x , v = − cos x ; {\displaystyle du=dx,\;\;v=-\cos x;} todėl
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} apskaičiavome, du kartus pritaikę dalinio integravimo metodą. Nesunku suvokti, kad integralą ∫ 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.} Gausime 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.} Gausime 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}}.} Jis apskaičiuojamas pagal ∫ u d v = u v − ∫ v d u {\displaystyle \int u\;{\mathsf {d}}v=uv-\int v\;{\mathsf {d}}u} formulę, tarus, kad u = x , d v = d x cos 2 x . {\displaystyle u=x,\;\;dv={\frac {dx}{\cos ^{2}x}}.} Tada 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.} Tada 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.} Darome keitinį t = x . {\displaystyle t={\sqrt {x}}.} Tada t 2 = x , 2 t d t = d x . {\displaystyle t^{2}=x,\;\;2t\;dt=dx.} Tuomet turime:∫ 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}}.} Tada
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.} Išintegruosime tokį integralą dalimis iš karto.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}.} Tada
∫ 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.}