Question

Evaluate the definite integral ${\int }_2^3\frac{dx}{x^2-1}$

Answer 1

Let $I={\int }_2^3\frac{dx}{x^2-1}$
$\Rightarrow \int \frac{dx}{x^2-1}=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|=F\left(x\right)$
By second fundamental theorem of calculus, we obtain
$I=F\left(3\right)-F\left(2\right)$
$=\frac{1}{2}[\log \left|\frac{3-1}{3+1}\right|-\log \left|\frac{2-1}{2+1}\right|]$
$=\frac{1}{2}[\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|]$
$=\frac{1}{2}[\log \frac{1}{2}-\log \frac{1}{3}]$
$=\frac{1}{2}\left[\log \frac{3}{2}\right]$