# Question #61a6c

Apr 12, 2017

Always let "u" equal the polynomial because differentiating it makes the resulting integral simpler

#### Explanation:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

let $u = {x}^{2} + 2 x + 1$, then $\mathrm{du} = 2 x + 2 \mathrm{dx}$
let $\mathrm{dv} = {e}^{7 x} \mathrm{dx}$, then $v = \frac{1}{7} {e}^{7 x}$

$\int \left({x}^{2} + 2 x + 1\right) {e}^{7 x} \mathrm{dx} = \frac{1}{7} \left({x}^{2} + 2 x + 1\right) {e}^{7 x} - \frac{1}{7} \int \left(2 x + 2\right) {e}^{7 x} \mathrm{dx}$

Repeat the integration by parts:

$\int \left({x}^{2} + 2 x + 1\right) {e}^{7 x} \mathrm{dx} = \frac{1}{7} \left({x}^{2} + 2 x + 1\right) {e}^{7 x} - \frac{1}{7} \int \left(2 x + 2\right) {e}^{7 x} \mathrm{dx}$

let $u = \frac{1}{7} \left(2 x + 2\right)$, then $\mathrm{du} = 2 \mathrm{dx}$
let $\mathrm{dv} = {e}^{7 x} \mathrm{dx}$, then $v = \frac{1}{7} {e}^{7 x}$

$\int \left({x}^{2} + 2 x + 1\right) {e}^{7 x} \mathrm{dx} = \frac{1}{7} \left({x}^{2} + 2 x + 1\right) {e}^{7 x} - \frac{1}{49} \left(2 x + 2\right) {e}^{7 x} + \frac{2}{49} \int {e}^{7 x} \mathrm{dx}$

$\int \left({x}^{2} + 2 x + 1\right) {e}^{7 x} \mathrm{dx} = \frac{1}{7} \left({x}^{2} + 2 x + 1\right) {e}^{7 x} - \frac{1}{49} \left(2 x + 2\right) {e}^{7 x} + \frac{2}{343} {e}^{7 x} + C$