Q. $\int_{}^{}{\left( \sqrt[3]{x} \right)\left( \sqrt[5]{1 + \sqrt[3]{x^{4}}} \right)}dx$ is equal to
A) $\left( 1 + x^{\frac{3}{4}} \right)^{\frac{6}{5}} + c$
B) $\left( 1 + x^{\frac{4}{3}} \right)^{\frac{6}{5}} + c$
C) $\frac{5}{8}\left( 1 + x^{\frac{4}{3}} \right)^{\frac{6}{5}} + c$
D) $\frac{1}{6}\left( 1 + x^{\frac{4}{3}} \right)^{6} + c$
Explanation
(c)
Let $I = \int_{}^{}{\left( \sqrt[3]{x} \right)\left( \sqrt[5]{1 + \sqrt[3]{x^{4}}} \right)}dx$
Put $\sqrt[3]{x^{4}} = t$
$\Rightarrow \ \frac{4}{3}\ .\sqrt[3]{x}\ dx = dt$
$\therefore\ I = \frac{3}{4}\int_{}^{}\left( \sqrt[5]{1 + t} \right)\ dt$
$I = \frac{3}{4}\left\lbrack \frac{(1 + t)^{\frac{1}{5} + 1}}{\frac{1}{5} + 1} \right\rbrack + c$
$= \frac{5}{8}\lbrack\left( 1 + \sqrt[3]{x^{4}} \right)^{6/5}\rbrack + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.