9x^2+6x+5 = 9x^2+6x+1 + 4 = (3x+1)^2+ 2^2 ∫1(3x+1)^2+2^2 dx 16tan^ 13x+12 + C ...

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Integrating Factor

∫19x2+6x+5dx.

Question

$\int \frac{1}{9 x^{2} + 6 x + 5} d x$ .

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Solution

$9 x^{2} + 6 x + 5 = 9 x^{2} + 6 x + 1 + 4 = (3 x + 1)^{2} + 2^{2}$
$\int \frac{1}{(3 x + 1)^{2} + 2^{2}} d x$
$\frac{1}{6} {tan}^{- 1} \frac{3 x + 1}{2} + C$

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