Question

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

• A. ${10}^x-x^{10}+C$
• B. ${10}^x+x^{10}+C$
• C. $({10}^x-x^{10}{)}^{-1}+C$
• D. $\log ({10}^x+x^{10})+C$

Answer 1

The correct option is D $\log ({10}^x+x^{10})+C$

Let $x^{10}+{10}^x=t$

$\therefore (10x^9+{10}^x{\log }_{e}10)dx=dt$

$\Rightarrow \int \frac{10x^9+{10}^x{\log }_{e}10}{x^{10}+{10}^x}dx=\int \frac{dt}{t}$

$=\log t+C$

$=\log ({10}^x+x^{10})+C$