Question

$\text{If}\int \frac{\sqrt[3]{31-x^3}}{x^5}dx=A\left(x\right){\left(\sqrt[3]{31-x^3}\right)}^m+C,$ where $A\left(x\right)$ is a function of $x$ and $C$ is a constant, then the value of $m$ is equal to

Answer 1

$\text{Let}I=\int \frac{\sqrt[3]{31-x^3}}{x^5}dx=\int \frac{\sqrt[3]{3\frac{1}{x^3}-1}}{x^4}dx\text{Put}\frac{1}{x^3}-1=t\text{Then}I=\frac{-1}{3}\int t^{\frac{1}{3}}dt=\frac{-1}{4x^4}{\left(\sqrt[3]{3(1-x^3}\right)}^4+C\therefore m=4$