$\int\limits_0^1 {\cfrac{{dx}}{{1 + {x^2}}}}$

: $\int\limits_0^1 {\cfrac{{dx}}{{1 + {x^2}}}} = \left[ {{{\tan }^{ - 1}}x} \right]_0^1 = {\tan ^{ - 1}}\left( 1 \right) - {\tan ^{ - 1}}\left( 0 \right) = \cfrac{\pi }{4}$