$$\int u(x)v'(x)dx=u(x)v(x)-\int u'(x)v(x)dx $$
Zad.1
$$ \int x e^x dx=||u=x , u’=1, v’=e^x,v=e^x|| = x e^x – \int e^x dx = x e^x – e^x + C $$
Zad.2
$$ \int x^2 \ln x dx =||u=lnx , u’=\frac{1}{x}, v’=x^2,v=\frac{1}{3}x^3||= \frac{x^3}{3} \ln x – \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \ln x – \frac{x^3}{9} + C $$
Zad.3
$$ \int \arctan x dx=||u=arctanx , u’= \frac{1}{1+x^2},v’=1,v=x|| = x \arctan x – \int x \cdot \frac{1}{1+x^2} , dx = x \arctan x – \frac{1}{2} \ln(1+x^2) + C $$
Zad.4
$$ \int x \sin x dx =||u=x , u’=1, v’=e^x,v=e^x|| = -x \cos x + \int \cos x , dx = -x \cos x + \sin x + C $$
Zad.5
$$ \int x^2 e^{x} dx=||u=x^2 , u’=2x, v’=e^x,v=e^x||=x^2e^x-\int 2x e^{x} dx =x^2e^x- 2x e^x +2 e^x +C $$
W rozwiązaniu wykorzystaliśmy wynik całki z zad.1
Zad.6
$$ \int \ln(x^2 + 1) dx=||u=ln(x^2 + 1) , u’=\frac{2x}{x^2+1}1, v’=1,v=x|| =x \ln(x^2 + 1) – \int \frac{2x^2}{x^2 + 1} dx = x \ln(x^2 + 1) – 2x + 2 \arctan x + C $$ $$\int \frac{2x^2}{x^2 + 1} dx = \int \frac{2x^2+2-2}{x^2 + 1} dx =\int \frac{2x^2+2}{x^2 + 1} dx -\int \frac{2}{x^2 + 1} dx =\int2dx-2\int \frac{dx}{x^2 + 1} =2x -2arctanx +C$$
Zad.7
$$ \int e^x \sin x dx=||u=sinx , u’=cosx, v’=e^x, v=e^x||= e^xsinx-\int e^x cosx= \frac{1}{2} e^x (\sin x – \cos x) + C $$ $$ \int e^x cosxdx=||u=cosx , u’=-sinx, v’=e^x, v=e^x||=e^xcosx+\int e^x sinxdx $$
Zad.8
$$ \int x \arcsin x dx =||u=arsinx, u’= \frac{dx}{\sqrt{1-x^2}} ,v”=x, v=\frac{1}{2x^2}||= x \sqrt{1-x^2} + \arcsin x – \int \frac{dx}{\sqrt{1-x^2}}$$ $$ = x \sqrt{1-x^2} + \arcsin x – \ln|x + \sqrt{1-x^2}| + C $$
Zad.9
$$ \int x^3 \cos x dx = =||u=x^3 , u’=3x^2, v’=cosx,v=sinx|| = -x^3sinx-\int -3x^2 \sin x dx = x^3 \sin x – 3 \int x^2 \sin x dx $$
$$ 3\int x^2 \sin x dx =||u=x^2 , u’=2x, v’=sinx,v=-cosx||= -3x^2 \sin x +3 \int 2x \cos x dx =-3x^2 \sin x +6 \int x \cos xdx $$ $$6 \int x \cos xdx=||u=x , u’=1, v’=sinx,v=-cosx||= -6xcosx +6 \int cos xdx=-6xcosx +6 sin x+ C $$ $$ \int x^3 \cos x dx = x^3 \sin x+3x^2 sinx +6xcosx -6 sin x+ C $$
Zad.10
$$ \int \ln(\sin x) dx =||u=ln(sinx) , u’=\frac{cosx}{sinx}, v’=1,v=x||= x \ln(\sin x) – \int \frac{x \cos x}{\sin x} dx=x \ln(\sin x) – \int xctgx dx $$ $$\int xctgx dx =||u=x , u’=1, v’=ctgx,v=ln(sinx)||=xln(sinx) -\int ln(sinx)dx$$ Zadanie wymaga innych technik całkowania.
Zad.11
$$ \int e^{2x} cos(3x)dx= ||u=e^{2x} , u’=2e^{2x}, v’=cos3x,v=\frac{sin3x}{3}||=\frac{e^{2x}sin3x}{3}-\int \frac{2e^{2x}sin3x}{3}dx $$$$\int \frac{2e^{2x}sin3x}{3}dx =||u=\frac{2e^{2x}}{3} , u’=\frac{4e^{2x}}{3}, v’=sin(3x),v=\frac{-cos(3x)}{3}||=\frac{-2cos(3x)e^{2x}}{9}- \int \frac{-4cos(3x)e^{2x}}{9}dx $$ $$ \int e^{2x} cos(3x)dx=
\frac{e^{2x}sin3x}{3} + \frac{2cos(3x)e^{2x}}{9}- \int \frac{4cos(3x)e^{2x}}{9}dx $$
$$\int e^{2x} cos(3x)dx= \frac{e^{2x}}{13} (2 \cos(3x) + 3 \sin(3x)) + C $$
Zad.12
$$ \int x^2 \arctan x dx= ||u=arctanx , u’=\frac{1}{x^2+1}, v’=x^2,v=\frac{x^3}{3}||= \frac{x^3}{3} \arctan x – \int \frac{x^3}{3(1+x^2)} dx $$ $$\int \frac{x^3}{3(1+x^2)} dx=\int \frac{x^3+x-x}{3(1+x^2)} dx=\int \frac{x(x^2+1}{3(1+x^2)}-\frac{x}{3(1+x^2)} dx=\int\frac{x}{3}dx-\int\frac{2x}{6(1+x^2)}dx= \frac{x^2}{6} + \frac{ln(1+x^2)}{6} +C$$
Zad.13
$$ \int \frac{\ln x}{x} dx=||u=lnx ,u’=\frac{1}{x},v’=\frac{1}{x},v=lnx||=(lnx)^2-\int \frac{\ln x}{x} dx = \frac{1}{2} (\ln x)^2 + C $$
Zad.14
$$ \int \sin^2(x) dx =||u=sin^2x, u’=2sinx cosx=sin2x, v’=1, v=x||= xsin^2x -\int xsin2x dx $$ $$\int xsin2x dx =||u=x, u’=1 ,v’=sin2x ,v=\frac{-cos2x}{2}||= \frac{-xcos2x}{2}- \int \frac{-cos2x}{2} = \frac{-xcos2x}{2}+ \frac{sin2x}{4} +C $$
$$ \int \sin^2(x) dx =xsin^2x+ \frac{xcos2x}{2}-\frac{sin2x}{4} +C $$