The value of $\int\frac{(x^4-x)^{1/4}}{x^5}dx$ is equal to (where C is an arbitrary constant)

$\frac{4}{15}(1-\frac{1}{x^3})^{5/4}+C$

The correct answer is Option (2) → $\frac{4}{15}(1+\frac{1}{x^3})^{5/4}+C$ **

$\displaystyle \int\frac{(x^{4}-x)^{1/4}}{x^{5}}\,dx$

Put $u=1-\frac{1}{x^{3}}\Rightarrow du=3x^{-4}\,dx\Rightarrow x^{-4}\,dx=\frac{du}{3}$

Since $x^{4}-x=x(x^{3}-1)$, $(x^{4}-x)^{1/4}=x^{1/4}(x^{3}-1)^{1/4}$ and with $x^{3}-1=x^{3}u$ the integrand becomes $x^{-4}u^{1/4}$.

$\displaystyle \int\frac{(x^{4}-x)^{1/4}}{x^{5}}\,dx=\frac{1}{3}\int u^{1/4}\,du=\frac{1}{3}\cdot\frac{u^{5/4}}{5/4}+C$

$\displaystyle =\frac{4}{15}\left(1-\frac{1}{x^{3}}\right)^{5/4}+C$