Question

Evaluate the integrals
${\int }_3^4\frac{1}{(x^2-4)}dx$

Answer 1

${\int }_3^4\frac{1}{(x^2-4)}dx={\int }_3^4\frac{1}{(x-2)(x+2)}dx$

$={\int }_3^4\frac{1}{4}\left[\frac{(x+2)-(x-2)}{(x-2)(x+2)}\right]dx={\int }_3^4\frac{1}{4}[\frac{1}{x-2}-\frac{1}{x+2}]dx$

$\Rightarrow \frac{1}{4}{\left|\ln \left(\frac{x-2}{x+2}\right)\right|}_3^4=\frac{1}{4}\ln \left(\frac{5}{3}\right)$