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