Q. The value of $\int_{}^{}{e^{\tan^{- 1}x}.\frac{(1 + x + x^{2})}{1 + x^{2}}\ dx}$ is
A) $\tan^{- 1}{x + c}$
B) $e^{\tan^{- 1}x} + 2x + c$
C) $e^{\tan^{- 1}x} + c$
D) $xe^{\tan^{- 1}x} + c$
Explanation
(d)
Let $I = \int_{}^{}{e^{\tan^{- 1}x}dx + \int_{}^{}e^{\tan^{- 1}x}.\frac{x}{\left( 1 + x^{2} \right)}\ dx}$
$= \int_{}^{}\frac{d}{dx}\left( x\ e^{\tan^{- 1}x} \right)dx + c$
$= xe^{\tan^{- 1}x} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.