Q. $\int_{}^{}{\frac{x^{2}}{1 + x^{6}}dx}$ is equal to
A) $x^{3} + c$
B) $\frac{1}{3}\tan^{- 1}{\left( x^{3} \right) + c}$
C) $\log{\left( 1 + x^{3} \right) + c}$
D) None of these
Explanation
(b)
Let $I = \int_{}^{}{\frac{x^{2}}{1 + \left( x^{3} \right)^{2}}dx}$
Put $x^{3} = t\ \Rightarrow 3x^{2}dx = dt$
$\therefore I = \frac{1}{3}\int_{}^{}\frac{dt}{1 + t^{2}}$
$= \frac{1}{3}\tan^{- 1}{t + c = \frac{1}{3}\tan^{- 1}{\left( x^{3} \right) + c}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.