Question

Value of $\int \frac{\left(x-x^{3}\right)^{1 / 3}}{x^{4}} d x$ is

• 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. None of these

Answer 1

Answer: (B)

$\int \frac{\left(x-x^{3}\right)^{1 / 3}}{x^{4}} d x$
$=\int \frac{1}{x^{3}}\left(\frac{1}{x^{2}}-1\right)^{1 / 3} d x$
$=\frac{-1}{2} \int t ^{1 / 3} dt \left[\right.$ Putting $\left.\frac{1}{ x ^{2}}-1= t\Rightarrow \frac{-2}{ x ^{3}} d x = dt \right]$
$=\frac{-1}{2} \cdot \frac{t^{4 / 3}}{4 / 3}+C=\frac{-3}{8}\left(\frac{1}{x^{2}}-1\right)^{4 / 3}+C$