<s>[INST] Partial fraction integration So I want to find all antiderivaties of
$\frac{x}{x^3-1}$
Since the denominator is of a lesser degree than the numerator, partial fraction is to be used instead of long division.
I've started by doing:
$\frac{x}{x^3-1} = \frac{A}{x^2+x+1} + \frac{B}{x-1}$
Hence,
$x = A(x-1)+B(x^2+x+1)$
Then setting $x = 1$ to get $3B = 1 => B = \frac{1}{3}$
So I have $\int({\frac{A}{x^2+x+1}}+{\frac{1}{3(x-1)}})$
However, I'm not sure how to continue from here.
 [/INST] You can do partial fractions in this way: $$\frac{x}{x^{3}-1} = \frac{Ax+B}{x^{2}+x+1} + \frac{C}{x-1}$$
Now by equating the coefficients you get:


*

*$A+C=0$

*$B+C-A=1$ and;

*$C-B=0$. 
Solving you get the values of $A= -\frac{1}{3}$, $B=\frac{1}{3}$ and $C=\frac{1}{3}$.
So, the required integral breaks down as $$ \int\frac{x}{x^{3}-1} \ dx = -\frac{1}{3}\int \frac{x -1}{x^{2}+x+1} \ dx + \frac{1}{3} \int \frac{1}{x-1} \ dx$$
Now $$\int\frac{x-1}{x^{2}+x+1} \ dx = \frac{1}{2} \int \frac{2x-2}{x^{2}+x+1} \ dx =\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1}  -\frac{1}{2} \int \frac{3}{x^{2}+x+1} \ dx$$
The first part of the integral is easy. For the second part use $$x^{2}+x+1 = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \Bigl(\frac{\sqrt{3}}{2}\Bigr)^{2}$$ and then put $t=x+\frac{1}{2}$ and then the trignometric substitution.
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