# How do you find the antiderivative of x^2*e^(2x)?

Jan 23, 2017

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

#### Explanation:

$\int {x}^{2} {e}^{2 x} \mathrm{dx} = \int {x}^{2} {\left(\frac{1}{2} {e}^{2 x}\right)}^{p} r i m e \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \int {\left({x}^{2}\right)}^{p} r i m e \cdot \frac{1}{2} {e}^{2 x} \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \int x {e}^{2 x} \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \int x {\left(\frac{1}{2} {e}^{2 x}\right)}^{p} r i m e \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \left(\frac{1}{2} x {e}^{2 x} - \int {\left(x\right)}^{p} r i m e \cdot \frac{1}{2} {e}^{2 x} \mathrm{dx}\right)$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \frac{1}{2} x {e}^{2 x} + \int \frac{1}{2} {e}^{2 x} \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \frac{1}{2} x {e}^{2 x} + \frac{1}{4} {e}^{2 x} + C$

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