$I=\int{\frac{\left( 1-{{x}^{5}} \right){{x}^{4}}dx}{{{x}^{5}}\left( 1+{{x}^{5}} \right)}}=\frac{1}{5}\int{\frac{\left( 1-{{x}^{5}} \right)d\left( {{x}^{5}} \right)}{{{x}^{5}}\left( 1+{{x}^{5}} \right)}}=\frac{1}{5}\int{\left( \frac{1}{{{x}^{5}}}-\frac{2}{1+{{x}^{5}}} \right)}d\left( {{x}^{5}} \right)=\frac{1}{5}\left[ \ln \left| {{x}^{5}} \right|-2\ln \left| 1+{{x}^{5}} \right| \right]+C$

Suy ra: $a=\frac{1}{5};b=-2\Rightarrow 10a+b-0$