Prove

$\int \frac{1}{9 x^{2}+6 x+5} d x=\int \frac{1}{(3 x+1)^{2}+2^{2}} d x$

Let $(3 x+1)=t$

$\therefore 3 d x=d t$

$\Rightarrow \int \frac{1}{(3 x+1)^{2}+2^{2}} d x=\frac{1}{3} \int \frac{1}{t^{2}+2^{2}} d t$

$=\frac{1}{3 \times 2} \tan ^{-1} \frac{t}{2}+C$

$=\frac{1}{6} \tan ^{-1}\left(\frac{3 x+1}{2}\right)+C$