Evaluate : 1 x^2+8 x+12 d x - Vidyadip

Question

Evaluate : $\int \frac{1}{x^2+8 x+12} \cdot d x$

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Answer

$
\begin{aligned}
\mathrm{I} & =\int \frac{1}{x^2+8 x+16-4} \cdot d x \\
& =\int \frac{1}{(x+4)^2-(2)^2} \cdot d x
\end{aligned}
$
$
\begin{array}{r}
\because \int \frac{1}{x^2-a^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)+c \\
\mathrm{I}=\frac{1}{2(2)} \cdot \log \left(\frac{(x+4)-2}{(x+4)+2}\right)+c \\
=\frac{1}{4} \cdot \log \left(\frac{x+2}{x+6}\right)+c \\
\therefore \quad \int \frac{1}{x^2+8 x+12} \cdot d x=\frac{1}{4} \cdot \log \left(\frac{x+2}{x+6}\right)+c
\end{array}
$

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