Evaluate $\int\frac{\sqrt[4]{x^4-x}}{x^5}\:\:dx$

Question

Evaluate $$\int\frac{\sqrt[4]{x^4-x}}{x^5}\:\:dx$$

I tried to do the $u$ substituion but every time I come to a very complicated expression. Like

Let $$\frac{x^4-x}{x^{20}}=u^4$$ and the and then differentiating it. It leads to a very intimidating expression. How can I do it other way$?$ Thanks.

Answer 1

You were right about the $u$ substitution idea. But the implementation was not correct.

Let $u=\frac{1}{x^3}$ then $dx=du\cdot\frac{x^4}{-3}$ and the integral becomes $$\begin{align} \int\frac{\sqrt[4]{x^4-x}}{x^5}\cdot\frac{x^4}{-3}\cdot du &=\frac{-1}{3}\int\sqrt[4]{\frac{x^4-x}{x^4}}\:\:du\\ &=\frac{-1}{3}\int\sqrt[4]{1-u}\:\:du \end{align} $$ I hope you can carry on now.

Answer 2

Hint:

$$\int\frac{\sqrt[4]{x^4-x}}{x^5}\ dx=\int\frac1{x^4}\sqrt[4]{1-\frac1{x^3}}\ dx$$

Now try a $u=-\frac1{x^3}$ sub.