Consider, $$I=\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}}dx$$
Let, $$\tan^{-1}x=t$$.
Then, $$d(\tan^{-1}x)=dt\Rightarrow dx=(1+x^{2})dt$$
Also, $$x=0\Rightarrow t=\tan^{-1}0=0$$ and $$x=1\Rightarrow t=\tan^{-1}1=\dfrac{\pi}{4}$$
$$\therefore I=\displaystyle\int_{0}^{\pi/4}t\ dt$$
$$I=\left[\dfrac{t^{2}}{2}\right]_{0}^{\pi/4}$$
$$I=\dfrac{\pi^{2}}{32}$$