Question

$\int \frac{\left(x^{4}-x\right)^{1 / 4}}{x^{5}} d x$ 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)

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