Official solution
\[\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + \,\,...\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)\]
\[ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{n^2}} \over {({n^2} + {r^2})(n + r)}}} \]
\[ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{1 \over {\left[ {1 + {{\left( {{r \over n}} \right)}^2}} \right]\left[ {1 + \left( {{r \over n}} \right)} \right]}}} \]
\[ = \int\limits_0^1 {{1 \over {(1 + {x^2})(1 + x)}}dx} \]
\[ = {1 \over 2}\int_0^1 {\left[ {{1 \over {1 + x}} - {{(x - 1)} \over {(1 + {x^2})}}} \right]dx} \]
\[ = {1 \over 2}\left[ {\ln (1 + x) - {1 \over 2}\ln \left( {1 + {x^2}} \right) + {{\tan }^{ - 1}}x} \right]_0^1\]
\[ = {1 \over 2}\left[ {{\pi \over 4} + {1 \over 2}\ln 2} \right] = {\pi \over 8} + {1 \over 4}\ln 2\]