Q. If $I = \int_{}^{}\frac{x^{5}}{\sqrt{1 + x^{3}}}\ dx$, then $I$ is equal to
A) $\frac{2}{9}\left( 1 + x^{3} \right)^{\frac{5}{2}} + \frac{2}{3}\left( 1 + x^{3} \right)^{\frac{3}{2}} + c$
B) $\log\left| \sqrt{x} + \sqrt{1 + x^{3}} \right| + c$
C) $\log\left| \sqrt{x} - \sqrt{1 - x^{3}} \right| + c$
D) $\frac{2}{9}\left( 1 + x^{3} \right)^{\frac{3}{2}} - \frac{2}{3}\left( 1 + x^{3} \right)^{\frac{1}{2}} + c$
Explanation
(d)
Given, $I = \int_{}^{}\frac{x^{5}}{\sqrt{1 + x^{3}}}dx$
Let $1 + x^{3} = t\ \Rightarrow \ 3x^{2}dx = dt$
$\therefore I = \int_{}^{}\frac{(t - 1)}{\sqrt{t}}.\frac{dt}{3} = \frac{1}{3}\int_{}^{}\left( \sqrt{t} - t^{\frac{- 1}{2}} \right)dt$
$= \frac{1}{3}\left\lbrack \frac{2t^{\frac{3}{2}}}{3} - 2t^{\frac{1}{2}} \right\rbrack + c$
$= \frac{2}{9}\left( 1 + x^{3} \right)^{\frac{3}{2}} - \frac{2}{3}\left( 1 + x^{3} \right)^{\frac{1}{2}} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.