Q. $\int_{}^{}{\frac{({\tan^{- 1}{x)}}^{3}}{1 + x^{2}}\ dx}$ is equal to
A) $3\ ({\tan^{- 1}{x)}}^{2} + c$
B) $\frac{({\tan^{- 1}{x)}}^{4}}{4} + c$
C) $({\tan^{- 1}{x)}}^{4} + c$
D) None of these
Explanation
(b)
Let $\tan^{- 1}{x = t}$
$\Rightarrow \ \frac{1}{1 + x^{2}}dx = dt$
$\therefore\ \int_{}^{}{\frac{({\tan^{- 1}{x)}}^{3}}{1 + x^{2}}dx = \int_{}^{}{t^{3}dt}}$
$= \frac{t^{4}}{4} + c = \frac{({\tan^{- 1}{x)}}^{4}}{4} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.