$$\int\frac{dx}{x-\sqrt[4]{x}}$$
given the substitution $x=u^{4}, dx=4u^{3}du$
$$=\int\frac{4u^{3}du}{u^{4}-u}=\int\frac{4u^{3}du}{u(u^{3}-1)}=\int\frac{4u^{2}du}{(u^{3}-1)}$$
At this point I believe I can factor the denominator and use partial fractions
$$\int\frac{4u^{2}du}{(u-1)(u^{2}+u+1)}=\int\bigg(\frac{A}{u-1}+\frac{B}{u^2+u+1}\bigg)du$$
by substituting in $u=1$ and ignoring the factor in the denominator $(u-1)$ (Grubby Thumb Rule) $A=\frac{4*1^{2}}{1^{2}+1+1}, A=\frac{4}{3}$
and substituting arbitrarily $u=0$ into the expression $$4u^{2}=\frac{4}{3}(u^2+u+1)+B(u-1)$$ $$0=\frac{4}{3}(1)-B, B=\frac{4}{3}$$
therefore
$$\int\bigg(\frac{A}{u-1}+\frac{B}{u^2+u+1}\bigg)du=\int\bigg(\frac{4}{3(u-1)}+\frac{4}{3(u^2+u+1)}\bigg)du$$
I am able to solve this however I dont think the value for B is correct
I have the following questions:
1: Am I able to use partial fractions even though the denominator of B is $u^{2}+u+1$
2: Did I do the substitution properly?
3: am I able to do what I did to find the value for A?
Thanks Stax
If you have to use partial fractions, you need to start with $$\frac{4u^{2}}{(u-1)(u^{2}+u+1)}=\frac{A}{u-1}+\frac{Bu+C}{u^2+u+1}$$ and then follow pretty much the procedure you have tried.
The easy way to evaluate $$\int\frac{4u^{2}du}{u^{3}-1}=\frac43\int\frac{3u^{2}du}{u^{3}-1}$$ is by substituting $v=u^3-1$, which gives the answer immediately.