Evaluate the following definite integrals : _1^2 d x x^2+6 x+5 - Vidyadip

Question

Evaluate the following definite integrals : $\int_1^2 \frac{d x}{x^2+6 x+5}$

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Answer

$
\begin{aligned}
& =\int_1^2 \frac{d x}{\left(x^2+6 x+9\right)-4} \frac{d x}{x^2+6 x+5} \\
& =\int_1^2 \frac{1}{(x+3)^2-(2)^2} d x \\
& =\frac{1}{2(2)}\left[\log \left|\frac{x+3-2}{x+3+2}\right|\right]_1^2=\frac{1}{4}\left[\log \left|\frac{x+1}{x+5}\right|\right]_1^2 \\
& =\frac{1}{4}\left[\log \frac{3}{7}-\log \frac{2}{6}\right] \\
& =\frac{1}{4} \log \left(\frac{3}{7} \times \frac{6}{2}\right)=\frac{1}{4} \log \left(\frac{9}{7}\right) .
\end{aligned}
$

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