Question

$\int \frac{1}{{(3x+1)}^2+3^2}dx$
$\frac{1}{3}\times \frac{1}{3}{tan}^{-1}\left(\frac{3x+1}{3}\right)+c$

Answer 1

$\int \frac{dx}{{(3x+1)}^2+3^2}$

Let $t=3x+1\Rightarrow dt=3dx$

$\Rightarrow dx=\frac{1}{3}dt$

$\Rightarrow \int \frac{dx}{{(3x+1)}^2+3^2}$

$=\frac{1}{3}\int \frac{dt}{t^2+3^2}$

$=\frac{1}{3}\frac{1}{3}{\tan }^{-1}\frac{t}{3}+c$ where $c$ is the constant of integration.

$=\frac{1}{9}{\tan }^{-1}\left(\frac{3x+1}{3}\right)+c$ where $t=3x+1$