I need to solve this problem: $$\int{6\over x^3-1}dx$$
Here is what I have so far: $$\int{6\over x^3-1}dx = 6*\int{1\over(x-1)(x^2+x+1)}dx$$
Next, $${1\over(x-1)(x^2+x+1)} = {A\over x-1} + {Bx+C\over x^2+x+1}$$
Continued, $${1\over3}*\int{1\over x-1}dx - {1\over 3}*\int{x+2\over x^2+x+1}dx$$
First off, is that last step correct (after equating and stuff)? Also, what would the next step be?
I know that $${1\over3}*\int{1\over x-1}dx = {ln\lvert x-1\rvert\over3}$$ At least I hope it does. I just can't figure out $$ -{1\over 3}*\int{x+2\over x^2+x+1}dx$$ There aren't any trig identities, u-sub obviously would not work here, I'm not sure about integration by parts, I can't decompose it since the bottom is not factorable. I'm guessing I have to write it differently by splitting it somehow, I just can't see how though. Any help? At least a hint?
Thank you!
Hint. $$\frac{x + 2}{x^2 + x + 1} = \frac{1}{2} \cdot \frac{2x + 1}{x^2 + x + 1} + \frac{1.5}{(x + \frac{1}{2})^2 + \frac{3}{4}}$$