Show that

$\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}=\int_{-1}^{1} \frac{d x}{\left(x^{2}+2 x+1\right)+4}=\int_{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}$

Let $x+1=t \Rightarrow d x=d t$

When $x=-1, t=0$ and when $x=1, t=2$

$\therefore \int_{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}=\int_{0}^{2} \frac{d t}{t^{2}+2^{2}}$

$=\left[\frac{1}{2} \tan ^{-1} \frac{t}{2}\right]_{0}^{2}$

$=\frac{1}{2} \tan ^{-1} 1-\frac{1}{2} \tan ^{-1} 0$

$=\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{8}$