Question

Integrate the function $\frac{1}{\sqrt{9x^2+6x+5}}$

Answer 1

$\int \frac{1}{\sqrt{9x^2+6x+5}}dx=\int \frac{1}{\sqrt{(3x+1{)}^2+(2{)}^2}}dx$
Let $(3x+1)=t$
$\therefore 3dx=dt$
$\Rightarrow \int \frac{1}{\sqrt{(3x+1{)}^2+(2{)}^2}}dx=\frac{1}{3}\int \frac{1}{\sqrt{t^2+2^2}}dt$
$=\frac{1}{3}\left[\frac{1}{2}{\tan }^{-1}\left(\frac{t}{2}\right)\right]+C$
$=\frac{1}{6}{\tan }^{-1}\left(\frac{3x+1}{2}\right)+C$