$\int \frac{dx}{9 x^{2} + 6 x + 5}$ is equal to

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

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

$\cot^{- 1} (\frac{3 x + 1}{2}) + C$

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

answer is D.

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Detailed Solution

$\int \frac{dx}{9 x^{2} + 6 x + 5} = \frac{1}{9} \int \frac{dx}{x^{2} + \frac{2}{3} x + \frac{5}{9}}$
$\begin{array}{l} = \frac{1}{9} \int \frac{dx}{x^{2} + 2 \cdot \frac{1}{3} \cdot x + {(\frac{1}{3})}^{2} + \frac{5}{9} - {(\frac{1}{3})}^{2}} \\ = \frac{1}{9} \int \frac{dx}{x^{2} + \frac{2}{3} x + {(\frac{1}{3})}^{2} + \frac{5}{9} - \frac{1}{9}} \\ = \frac{1}{9} \int \frac{dx}{{(x + \frac{1}{3})}^{2} + {(\frac{2}{3})}^{2}} \\ = \frac{1}{9} (\frac{1}{2 / 3}) \tan^{- 1} (\frac{x + \frac{1}{3}}{2 / 3}) + C \end{array}$
$[∵ \int \frac{dx}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a})]$
$= \frac{1}{6} \tan^{- 1} (\frac{3 x + 1}{2}) + C$