Question

Evaluate the integrals
${\int }_{-1}^1\frac{1}{(x^2+6x+8)}dx$

Answer 1

$\Rightarrow {\int }_{-1}^1\frac{1}{[{(x+3)}^2-1^2]}dx$

Formula : $\int \frac{1}{x^2-a^2}=\frac{1}{2a}ln\left(\frac{x-a}{x+a}\right)$

$\Rightarrow \frac{1}{2.1}\ln {\left(\left|\frac{x+2}{x+4}\right|\right)}_{-1}^1$

$\Rightarrow \frac{1}{2}\ln \left(\frac{9}{5}\right)$