Q. $\int_{}^{}{\frac{10x^{9} + 10^{x}\log_{e}10}{10^{x} + x^{10}}\ dx}$ is equal to
A) $10^{x} - x^{10} + c$
B) $10^{x} + x^{10} + c$
C) $\left( 10^{x} - x^{10} \right)^{- 1} + c$
D) $\log{\left( 10^{x} + x^{10} \right) + c}$
Explanation
(d)
Put $10^{x} + x^{10} = t$
$\therefore(10^{x}\log_{e}{10 + 10x^{9})dx = dt}$
$\therefore\ \int_{}^{}{\frac{10x^{9} + 10^{x}\log_{e}10}{10^{x} + x^{10}}dx = \int_{}^{}\frac{dt}{t}}$
$= \log_{e}{t + c}$
$= \log_{e}{\left( 10^{x} + x^{10} \right) + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.