Question

The value of $\int \frac{(x^4-x{)}^{\frac{1}{4}}}{x^5}dx$ is

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

Answer 1

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

$\int \frac{{(x^4-x)}^{\frac{1}{4}}}{x^5}dx=\int \frac{{(1-\frac{1}{x^3})}^{\frac{1}{4}}}{x^4}dx$

Let $1-\frac{1}{x^3}=t\Rightarrow \frac{3}{x^4}dx=dt$
Therefore,
$\frac{1}{3}\int \frac{(x^4-x{)}^{\frac{1}{4}}}{x^5}dx=\frac{1}{3}\int t^{\frac{1}{4}}dt=\frac{1}{3}\times \frac{4}{5}{t}^{\frac{5}{4}}+C=\frac{4}{15}{(1-\frac{1}{x^3})}^{\frac{5}{4}}+C$