# How do you find the integral of xe^(x/2)dx ?

May 26, 2015

By part

$u = x$
$\mathrm{du} = 1$

$\mathrm{dv} = {e}^{\frac{1}{2} x}$

$v = 2 {e}^{\frac{1}{2} x}$

$\implies 2 \left(\left[x {e}^{\frac{1}{2} x}\right] - \int {e}^{\frac{1}{2} x} \mathrm{dx}\right) + C$

$\implies 2 \left(\left[x {e}^{\frac{1}{2} x}\right] - 2 \left[{e}^{\frac{1}{2} x}\right]\right) + C$

$\implies 2 {e}^{\frac{1}{2} x} \left(x - 2\right) + C$