Question

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

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