Question

$\int \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^4} d x$ is equal to

• A. $\frac{3}{8}\left(\frac{1}{ x ^2}-1\right)^{\frac{4}{3}}+ C$
• B. $\frac{-3}{8}\left(\frac{1}{ x ^2}+1\right)^{\frac{4}{3}}+ C$
• C. $\frac{-3}{8}\left(\frac{1}{ x ^2}-1\right)^{\frac{4}{3}}+ C$
• D. $\frac{-3}{4}\left(1-\frac{1}{ x ^2}\right)^{\frac{4}{3}}+ C$

Answer 1

Answer: (C)

$\int \frac{\left(\frac{1}{x^2}-1\right)^{\frac{1}{3}}}{x^3} d x$
Put $\frac{1}{x^2}-1=t$
$\Rightarrow \int \frac{- t ^{\frac{1}{3}}}{2} dt =\frac{-3 t ^{\frac{4}{3}}}{8}+ C $
$\Rightarrow \frac{-3}{8}\left(\frac{1}{ x ^2}-1\right)^{\frac{4}{3}}+ C$