# How do you integrate (t^2)e^(4t)?

Mar 9, 2018

$\frac{{t}^{2} {e}^{4 t}}{4} - \frac{1}{2} \left(\frac{t {e}^{4 t}}{4} - {e}^{4 t} / 16\right) + C$

#### Explanation:

We have $\int {t}^{2} {e}^{4 t} \mathrm{dt}$.
According to Integration by Parts, $\int f \left(t\right) g \left(t\right) \mathrm{dt} = f \left(t\right) \int g \left(t\right) \mathrm{dt} - \int f ' \left(x\right) \left(\int g \left(t\right) \mathrm{dt}\right) \mathrm{dt}$

Here, $f \left(t\right) = {t}^{2}$ and $g \left(t\right) = {e}^{4 t}$. So we input:

${t}^{2} \int {e}^{4 t} \mathrm{dt} - \int \left({t}^{2}\right) ' \left(\int {e}^{4 t} \mathrm{dt}\right) \mathrm{dt}$

A logical route to take is to find $\int {e}^{4 t} \mathrm{dt}$

According to Integration by Substitution, $\int f \left(g \left(t\right)\right) g ' \left(t\right) \mathrm{dt} = \int f \left(u\right) \mathrm{du}$, where $u = g \left(t\right)$. We can write the above as:

$\frac{1}{4} \int {e}^{4 t} 4 \mathrm{dt}$

$\frac{1}{4} \int {e}^{u} \mathrm{du}$

$\frac{1}{4} {e}^{u}$

${e}^{4 t} / 4$. So we input:

$\frac{{t}^{2} {e}^{4 t}}{4} - \frac{1}{2} \int t {e}^{4 t} \mathrm{dt}$

Apply integration by parts for the integral:

$t \int {e}^{4 t} \mathrm{dt} - \int t ' \left(\int {e}^{4 t} \mathrm{dt}\right) \mathrm{dt}$

Since we know that $\int {e}^{4 t} \mathrm{dt}$ is:

$\frac{t {e}^{4 t}}{4} - \int {e}^{4 t} / 4 \mathrm{dt}$

$\frac{t {e}^{4 t}}{4} - \frac{1}{4} \cdot \frac{1}{4} {e}^{4 t}$

$\frac{t {e}^{4 t}}{4} - {e}^{4 t} / 16$

We can input this into our eariler calculations:

$\frac{{t}^{2} {e}^{4 t}}{4} - \frac{1}{2} \left(\frac{t {e}^{4 t}}{4} - {e}^{4 t} / 16\right)$

Add the constant of integration:

$\frac{{t}^{2} {e}^{4 t}}{4} - \frac{1}{2} \left(\frac{t {e}^{4 t}}{4} - {e}^{4 t} / 16\right) + C$