Integration by Parts

Integration by Parts

Recall: The product rule.

$dfrac{d}{dx}(uv)=udfrac{dv}{dx}+vdfrac{du}{dx}$

Integration by parts is the reverse of this.

${LARGE int}udfrac{dv}{dx}=uv-{LARGE int}vdfrac{du}{dx}$

You should choose $u$ to be something that differentiates nicely, so that you can evaluate the integral on the right-hand side.

Example: Find $int xsin(x)dx$

$u=x;;;dfrac{dv}{dx}=sin(x)$

$dfrac{du}{dx}=1;;;v=-cos(x)$

$begin{aligned}int xsin(x)dx&=-xcos(x)-int-cos(x)dx[1.2em]&=-xcos(x)+int cos(x)dx[1.2em]&=-xcos(x)+sin(x)+cend{aligned}$

Integral of the Natural Logarithm

One of the most important uses of integration by parts is that it gives us a way to find $int ln(x)dx$, by writng $ln(x)$ as
$1times ln(x)$.

$u=ln(x);;;dfrac{dv}{dx}=1$

$dfrac{du}{dx}=dfrac{1}{x};;;v=x$

$begin{aligned}int ln(x)dx&=xln(x)-int xdfrac{1}{x}dx[1.2em]&=xln(x)-int1dx[1.2em]&=xln(x)-x+c[1.2em]&=x(ln(x)-1)+cend{aligned}$

Integrating by Parts Multiple Times

Some integrals require you to integrate by parts multiple times to get a solution to.

Example: Find $int x^{2}e^{x}dx$

$u=x^{2};;;dfrac{dv}{dx}=e^{x}$

$dfrac{du}{dx}=2x;;;v=e^{x}$

$begin{aligned}int x^{2}e^{x}dx=x^{2}e^{x}-int2xe^{x}dxend{aligned}$

We must integrate by parts again the second integral.

$u=2x;;;dfrac{dv}{dx}=e^{x}$

$dfrac{du}{dx}=2;;;v=e^{x}$

$begin{aligned}int x^{2}e^{x}dx&=x^{2}e^{x}-2xe^{x}+int2e^{x}dx[1.2em]&=x^{2}e^{x}-2xe^{x}+2e^{x}+cend{aligned}$