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

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

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.

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

$$\int\frac{\sqrt{x^4-x}}{x^5}\ dx=\int\frac1{x^4}\sqrt{1-\frac1{x^3}}\ dx$$
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{x^4-x}}{x^5}\cdot\frac{x^4}{-3}\cdot du &=\frac{-1}{3}\int\sqrt{\frac{x^4-x}{x^4}}\:\:du\\ &=\frac{-1}{3}\int\sqrt{1-u}\:\:du \end{align} I hope you can carry on now.