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Integrate the function
$\frac{1}{9 x^2+6 x+5}$
$\frac{1}{9 x^2+6 x+5}$
Solution:
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Verified Answer
$\int \frac{d x}{9 x^2+6 x+5}=\frac{1}{9} \int \frac{d x}{\left(x+\frac{1}{3}\right)^2+\left(\frac{2}{3}\right)^2}$
$=\frac{1}{6} \tan ^{-1}\left(\frac{3 x+1}{2}\right)+C$.
$=\frac{1}{6} \tan ^{-1}\left(\frac{3 x+1}{2}\right)+C$.
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