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

Answer

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

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