Question

Simplify: $\int \frac{{(x^4-x)}^{1/4}}{x^5}dx$

• A. $\frac{2}{5}{(1-\frac{1}{x^3})}^{-5/4}+c$
• B. $\frac{4}{15}{(1-\frac{1}{x^3})}^{-5/4}+c$
• C. $\frac{4}{15}{(1-\frac{1}{x^3})}^{5/4}+c$
• D. $\frac{2}{5}{(1-\frac{1}{x^3})}^{5/4}+c$

Answer 1

The correct option is C $\frac{4}{15}{(1-\frac{1}{x^3})}^{5/4}+c$

Let $I=\int \frac{{(x^4-x)}^{1/4}}{x^5}dx$
$=\int \frac{1}{x^4}{(1-\frac{1}{x^3})}^{1/4}dx$
Put $1-\frac{1}{x^3}=t$
$\Rightarrow \frac{3}{x^4}dx=dt$
$I=\frac{1}{3}\int t^{1/4}dt=\frac{4}{15}{t}^{5/4}+c$
$=\frac{4}{15}{(1-\frac{1}{x^3})}^{5/4}+c$