Question

$\int \frac{{\left(x^4-x\right)}^{1/4}}{x^5}dx$ is equal to

• A. $\frac{4}{15}{\left(1-\frac{1}{x^3}\right)}^{5/4}+C$
• B. $\frac{4}{5}{\left(1-\frac{1}{x^3}\right)}^{5/4}+C$
• C. $\frac{4}{15}{\left(1+\frac{1}{x^3}\right)}^{5/4}+C$
• D. None of these

Answer 1

Answer: (A)

$\int \frac{{\left(x^4-x\right)}^{1/4}}{x^5}dx=\int \frac{1}{x^4}{\left(1-\frac{1}{x^3}\right)}^{1/4}dx$
Putting $1-\frac{1}{x^3}=t$, we get
$I=\frac{1}{3}\int t^{1/4}dt=\frac{4}{15}{t}^{5/4}+C$
$=\frac{4}{15}{\left(1-\frac{1}{x^3}\right)}^{5/4}+C$