Official solution
f'(x) = tan--1(secx + tanx)
\[ \Rightarrow \] f'(x) = \[{\tan ^{ - 1}}\left( {{{1 + \sin x} \over {\cos x}}} \right)\] = \[{\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)\]
= \[{\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right)\]
As \[ - {\pi \over 2} < x < {\pi \over 2}\]
\[ \Rightarrow \] \[0 < {\pi \over 4} + {x \over 2} < {\pi \over 2}\]
\[ \therefore \] f'(x) = \[{\pi \over 4} + {x \over 2}\]
\[ \Rightarrow \] f(x) = \[{\pi \over 4}x + {{{x^2}} \over 4} + c\]
\[ \because \] f(0) = 0 \[ \Rightarrow \] c = 0
\[ \therefore \] f(x) = \[{\pi \over 4}x + {{{x^2}} \over 4}\]
\[ \Rightarrow \] f(1) = \[{{{\pi + 1} \over 4}}\]