Official solution
\[\int {{{\sqrt x } \over {{{(1 + x)}^2}}}} dx(x > 0)\]
Put x = tan2\[\theta \] \[ \Rightarrow \] 2xdx = 2tan\[\theta \]sec2\[\theta \]d\[\theta \]
\[I = \int {{{2{{\tan }^2}\theta .{{\sec }^2}\theta } \over 2}} d\theta = \int {2{{\sin }^2}\theta d\theta = \int {(1 - \cos 2\theta )d\theta } } \]
\[ = \theta - {{\sin 2\theta } \over 2} + c\]
\[ \Rightarrow f(x) = \theta - {1 \over 2} \times {{2\tan \theta } \over {1 + {{\tan }^2}\theta }} + c\] \[f(x) = \theta - {{\tan \theta } \over {1 + {{\tan }^2}\theta }} + c = {\tan ^{ - 1}}\sqrt x - {{\sqrt x } \over {1 + x}} + c\]
Now \[f(3) - f(1) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {{\sqrt 3 } \over {1 + 3}} - {\tan ^{ - 1}}(1) + {1 \over 2}\]
\[ = {\pi \over 3} - {{\sqrt 3 } \over 4} - {\pi \over 4} + {1 \over 2}\]
\[ = {\pi \over 12} + {1 \over 2} - {{\sqrt 3 } \over 4}\]