Question

$\int \frac{10x^9+{10}^x{\log }_{e}10}{{10}^x+x^{10}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{10x^9+{10}^x{\log }_{e}10}{{10}^x+x^{10}}dx \\ \mathrm{Putting}{10}^x+x^{10}=t \\ \Rightarrow {10}^x{\log }_{e}10+10x^9=\frac{dt}{dx} \\ \Rightarrow \left({10}^x{\log }_{e}10+10x^9\right)dx=dt \\ \therefore I=\int \frac{1}{t}dt \\ =\ln \left|t\right|+\mathrm{C} \\ =\ln \left|{10}^x+x^{10}\right|+\mathrm{C}\left[\because t={10}^x+x^{10}\right] \end{gathered}$$