$I=\int\limits_{1}^{2}{\frac{1}{x{{\left( x+1 \right)}^{2}}}}dx=\int\limits_{1}^{2}{\frac{x+1-x}{x{{\left( x+1 \right)}^{2}}}}dx=\int\limits_{1}^{2}{\frac{1}{x{{\left( x+1 \right)}^{2}}}}dx-\int\limits_{1}^{2}{\frac{1}{{{\left( x+1 \right)}^{2}}}}dx$

Suy ra

$I=\int\limits_{1}^{2}{\left( \frac{1}{x}-\frac{1}{x+1} \right)}dx$$-\int\limits_{1}^{2}{{{\left( x+1 \right)}^{-2}}}dx\left( x+1 \right)$$=\ln \left| \frac{x}{x+1} \right|\left| \begin{align}  & 2 \\  & 1 \\ \end{align} \right.$$+{{\left( x+1 \right)}^{-1}}\left| \begin{align}  & 2 \\  & 1 \\ \end{align} \right.$$=\ln \frac{4}{3}-\frac{1}{6}$

$\Rightarrow a=\frac{4}{3},b=-\frac{1}{6}\Rightarrow S=1$