Question

Evaluate the integral ${\int }_{-1}^1\frac{dx}{x^2+2x+5}$ using substitution.

Answer 1

${\int }_{-1}^1\frac{dx}{x^2+2x+5}={\int }_{-1}^1\frac{dx}{(x^2+2x+1)+4}={\int }_{-1}^1\frac{dx}{(x+1{)}^2+(2{)}^2}$
Let $x+1=t\Rightarrow dx=dt$
When $x=-1,t=0$ and When $x=1,t=2$
$\therefore {\int }_{-1}^1\frac{dx}{(x+1{)}^2+(2{)}^2}={\int }_0^2\frac{dt}{t^2+2^2}$
$={\left[\frac{1}{2}{\tan }^{-1}\frac{t}{2}\right]}_0^2$
$=\frac{1}{2}{\tan }^{-1}1-\frac{1}{2}{\tan }^{-1}0$
$=\frac{1}{2}\left(\frac{\pi }{4}\right)=\frac{\pi }{8}$