Ta có
$(\arctan x)' = \dfrac{1}{\sqrt{1+x^2}}$
Do đó
$d(\arctan x) = \dfrac{dx}{\sqrt{1 + x^2}}$
Thay vào ta có
$I = \int_0^1 \arctan x d(\arctan x) = \int_0^{\frac{\pi}{4}} udu = \dfrac{u^2}{2} \bigg\vert_0^{\frac{\pi}{4}}$
$= \dfrac{\pi^2}{16.2} -0 = \dfrac{\pi^2}{32}$.
Vậy $I = \dfrac{\pi^2}{32}$.
