Question

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

• A. $\frac{1}{2}\log ({10}^x+x^{10}).$
• B. $\log \left({10}^x\right).$
• C. $\log ({10}^x+x^{10}).$
• D. $\log ({10}^x-x^{10}).$

Answer 1

Answer: (C)

$\log ({10}^x+x^{10}).$

Let $I=\int \frac{10x^9+{10}^x{\log }_{e}10}{{10}^x+x^{10}}dx$
Put ${10}^x+x^{10}=t\Rightarrow (10x^9+{10}^x{\log }_{e}10)dx=dt$
Therefore
$I=\int \frac{dt}{t}=\log t=\log ({10}^x+x^{10})$
Hence, option 'C' is correct.