$S=\int_{1}^{2}{\frac{1}{x\left( 1+{{x}^{3}} \right)}}dx=\int_{1}^{2}{\frac{{{x}^{2}}dx}{{{x}^{3}}\left( 1+x \right)}}=\frac{1}{3}\int_{1}^{2}{\frac{d\left( {{x}^{3}} \right)}{{{x}^{3}}\left( 1+{{x}^{3}} \right)}}=\frac{1}{3}\left[ \int_{1}^{2}{\frac{d\left( {{x}^{3}} \right)}{{{x}^{3}}}-\int_{1}^{2}{\frac{d\left( {{x}^{3}} \right)}{1+{{x}^{3}}}}} \right]=\frac{1}{3}\ln \frac{{{x}^{3}}}{1+{{x}^{3}}}\left| \begin{align}  & 2 \\  & 1 \\ \end{align} \right|=\frac{1}{3}\ln \frac{16}{9}$ (dvđt)