# How can I evaluate the integral intx^2e^(4x)dx?

Jan 31, 2015

I would use integration by parts where you have:

$\int f \left(x\right) \cdot g \left(x\right) \mathrm{dx} = F \left(x\right) \cdot g \left(x\right) - \int F \left(x\right) \cdot g ' \left(x\right) \mathrm{dx}$

Where:

$F \left(x\right) = \int f \left(x\right) \mathrm{dx}$

$g ' \left(x\right)$ is the derivative of $g \left(x\right)$

I'll choose:

$f \left(x\right) = {e}^{4 x}$ and
$g \left(x\right) = {x}^{2}$

Integrating you get:

$\int {x}^{2} {e}^{4 x} \mathrm{dx} = {x}^{2} {e}^{4 x} / 4 - \int 2 x {e}^{4 x} / 4 \mathrm{dx} =$
$= {x}^{2} {e}^{4 x} / 4 - \int x {e}^{4 x} / 2 \mathrm{dx} =$
$= {x}^{2} {e}^{4 x} / 4 - \frac{1}{2} \int x {e}^{4 x} \mathrm{dx} =$ and by parts again:
$= {x}^{2} {e}^{4 x} / 4 - \frac{1}{2} \left[x {e}^{4 x} / 4 - \int 1 {e}^{4 x} / 4 \mathrm{dx}\right] =$
$= {x}^{2} {e}^{4 x} / 4 - x {e}^{4 x} / 8 + {e}^{4 x} / 32 + c =$
$= {e}^{4 x} \left({x}^{2} / 4 - \frac{x}{8} + \frac{1}{32}\right) + c$