Show that

$\int_{0}^{1} \frac{x}{x^{2}+1} d x$

Let $x^{2}+1=t \Rightarrow 2 x d x=d t$

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

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

$=\frac{1}{2}[\log |t|]_{1}^{2}$

$=\frac{1}{2}[\log 2-\log 1]$

$=\frac{1}{2} \log 2$