Q. The value of $\int_{}^{}\frac{1}{\left\lbrack (x - 1)^{3}(x + 2)^{5} \right\rbrack^{\frac{1}{4}}}\ dx$ is
A) $\frac{4}{3}\left( \frac{x - 1}{x + 2} \right)^{\frac{1}{4}} + c$
B) $\frac{4}{3}\left( \frac{x + 1}{x + 2} \right)^{\frac{1}{4}} + c$
C) $\frac{4}{3}\left( \frac{x + 1}{x - 2} \right)^{\frac{1}{4}} + c$
D) $\frac{4}{3}\left( \frac{x - 1}{x - 2} \right)^{\frac{1}{4}} + c$
Explanation
(a)
Let $\frac{x - 1}{x + 2} = t\ \Rightarrow \ \frac{1}{(x + 2)^{2}}dx = \frac{1}{3}\ dt$
$\therefore\ \int_{}^{}\frac{1}{\left\lbrack (x - 1)^{3}(x + 2)^{5} \right\rbrack^{\frac{1}{4}}}\ dx = \frac{1}{3}\int_{}^{}t^{- 3/4}dt$
$= \frac{4}{3}\ t^{1/4} + c$
$= \frac{4}{3}\left( \frac{x - 1}{x + 2} \right)^{\frac{1}{4}} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.