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.

  • 1
    $\begingroup$ Many fairly simple functions lead to integrals having no elementary closed form. Have you a reason to think this one has such an antiderivative? $\endgroup$
    – coffeemath
    Aug 17 at 10:51
  • $\begingroup$ Have you tried the substitution $u=\frac{1}{x^3} $? $\endgroup$
    – Lai
    Aug 17 at 10:55

2 Answers 2



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

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


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.


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