<s>[INST] Substitution and Partial Fractions (Integration) $$\int\frac{dx}{x-\sqrt[4]{x}}$$
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
 [/INST] $$
\int\frac{4u^{2}du}{u^3-1} = \frac 4 3 \int \frac {dw} w, \text{ etc.}
$$
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