Question

The integral given equals to

$\int \left(\sqrt[3]{x}\right)\left(\sqrt[5]{1+\sqrt[3]{x^4}}\right)dx=$

• A. ${\left(1+x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}^{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}+C$
• B. ${\left(1+x^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}+C$
• C. $\frac{5}{8}{\left(1+x^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}+C$
• D. $\frac{1}{6}{\left(1+x^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^6+C$
• E. $\frac{15}{7}{\left(1+x^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}+C$

Answer 1

The correct option is C

$\frac{5}{8}{\left(1+x^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}+C$

Explanation for correct option

The given integral is $\int \left(\sqrt[3]{x}\right)\left(\sqrt[5]{1+\sqrt[3]{x^4}}\right)dx$

Let $1+\sqrt[3]{x^4}=z^5$

$$\begin{gathered} \frac{4}{3}{x}^{\frac{1}{3}}dx=dz5z^4 \\ \int \left(\sqrt[3]{x}\right)\left(\sqrt[5]{1+\sqrt[3]{x^4}}\right)dx \\ =\int z·3·z^{4}dz \\ =\frac{5·3}{4}\int z^{5}dz \\ =\frac{5·3}{4}\frac{z^6}{6}+C \\ =5\frac{z^6}{8}+C \\ =\frac{5}{8}{\left(1+\sqrt[3]{x^4}\right)}^{\frac{6}{5}}+C \end{gathered}$$

Hence, option C is correct.