Evaluate _1^2 1 x^2+6 x+5 d x - Vidyadip

Question

Evaluate $\int_1^2 \frac{1}{x^2+6 x+5} d x$

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Answer

$\text { Let } I =\int_1^2 \frac{1}{x^2+6 x+5}$
$=\int_1^2 \frac{ d x}{x^2+6 x+9-9+5}$
$=\int_1^2 \frac{ d x}{(x+3)^2-4}$
$=\int_1^2 \frac{ d x}{(x+3)^2-(2)^2}$
$=\frac{1}{2 \times 2}\left[\log \left|\frac{x+3-2}{x+3+2}\right|\right]_1^2$
$=\frac{1}{4}\left[\log \left|\frac{x+1}{x+5}\right|\right]_1^2$
$=\frac{1}{4}\left[\log \left(\frac{3}{7}\right)-\log \left(\frac{2}{6}\right)\right]$
$=\frac{1}{4} \log \left(\frac{3}{7} \times \frac{6}{2}\right)$
$\therefore I =\frac{1}{4} \log \left(\frac{9}{7}\right)$

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