# Question #a8dd2

Nov 8, 2016

$\int {x}^{3} {e}^{2 {x}^{2}} \mathrm{dx} = \frac{1}{8} {e}^{2 {x}^{2}} \left(2 {x}^{2} - 1\right) + C$

#### Explanation:

$I = \int {x}^{3} {e}^{2 {x}^{2}} \mathrm{dx}$

Before using integration by parts (IBP), we can make a simpler substitution. Let $t = 2 {x}^{2}$ so that $\mathrm{dt} = 4 x \mathrm{dx}$. This also implies that ${x}^{2} = \frac{1}{2} t$.

Modifying the integral:

$I = \frac{1}{4} \int {x}^{2} {e}^{2 {x}^{2}} \left(4 x \mathrm{dx}\right) = \frac{1}{4} \int \frac{1}{2} t {e}^{t} \mathrm{dt} = \frac{1}{8} \int t {e}^{t} \mathrm{dt}$

Now we should apply IBP. This integration technique takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. So, for $\int t {e}^{t} \mathrm{dt}$, we should let:

$\left\{\begin{matrix}u = t & \implies & \mathrm{du} = \mathrm{dt} \\ \mathrm{dv} = {e}^{t} \mathrm{dt} & \implies & v = {e}^{t}\end{matrix}\right.$

Recall to differentiate $u$ and integrate $\mathrm{dv}$.

Thus:

$I = \frac{1}{8} \left(t {e}^{t} - \int {e}^{t} \mathrm{dt}\right) = \frac{1}{8} \left(t {e}^{t} - {e}^{t}\right) = \frac{1}{8} {e}^{t} \left(t - 1\right)$

Since $t = 2 {x}^{2}$ this becomes:

$I = \frac{1}{8} {e}^{2 {x}^{2}} \left(2 {x}^{2} - 1\right) + C$