$\int {\left( {\cfrac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{x^{10}} + {{10}^x}}}} \right)} dx$equals

Option d is correct

Let $I = \int {\left( {\cfrac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{x^{10}} + {{10}^x}}}} \right)} dx$

Put ${x^{10}} + {10^x} = t$ $\Rightarrow$ $\left( {10{x^9} + {{\log }_e}10 \cdot {{10}^x}} \right)dx = dt$

$\Rightarrow$ $I = \int {\cfrac{{dt}}{t} = \log \left| t \right|} + C = \log \left( {{x^{10}} + {{10}^x}} \right) + C$