How to evaluate: $$\int \frac{\mathrm{dx}}{x^4[x(x^5-1)]^{1/3}}$$ I have done a substantial work on it:
Let $x^5z^3=x^5-1$. So $$x^5(z^3-1)=1\implies 5x^4(z^3-1)\mathrm{d}x+x^5(-3z^2\mathrm{d}z)=0\implies \mathrm{d}x=\frac{3xz^2\mathrm{d}z}{5(z^3-1)}$$ So: $$\int \frac{\mathrm{d}x}{x^4[x(x^5-1)]^{1/3}}\text{ or }\int x^{-13/3}(x^5-1)^{-1/3}\mathrm{d}x\\ =\int x^{-13/3}(x^5z^3)^{-1/3}.\frac{3xz^2\mathrm{d}z}{5(z^3-1)}=\frac35\int \frac{x^{-13/3}x^{-5/3}z^{-1}xz^2\mathrm{d}z}{x^{-5}}\\ =\frac35\int z\;\mathrm{d}z=\frac{3}{10}\left(\frac{x^5-1}{x^5}\right)^{2/3}+\mathcal{C}$$ Is my working correct? Is there an easier way?
Actually(thanks to @Jean-ClaudeArbaut):
- $x^5(\color{red}{1-z^3})=1$
- $5x^4(z^3-1)\mathrm{d}x+x^5(\color{red}{3}z^2\mathrm{d}z)=0$
$$ \int \dfrac{1}{x^4\left[x\left(x^5-1\right)\right]^{1/3}}dx\tag{1} $$
use the sub $u = 1/x^3\implies du = -\dfrac{3}{x^4} dx$.
thus Eq. (1) becomes
$$ \int \dfrac{1}{\left(\dfrac{1}{u^2}-\dfrac{1}{u^{1/3}}\right)^{1/3}}\dfrac{du}{-3} = \dfrac{1}{-3}\int \dfrac{u^{2/3}}{\left(1-u^{5/3}\right)^{1/3}}du $$
then $v = \left(1-u^{5/3}\right)^{2/3}$
$$ dv = \dfrac{2}{3}\dfrac{-\dfrac{5}{3}u^{2/3}}{\left(1-u^{5/3}\right)^{1/3}}\rightarrow -\dfrac{9}{10}dv = \dfrac{u^{2/3}}{\left(1-u^{5/3}\right)^{1/3}}du $$
I think this correct, but then again I have made mistakes in the past!