# Integrate: $\int{6\over x^3-1}dx$

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!

• The \begin{align} · · · \end{align} environment and \cdot operator seem relevant here. You can read about them here. – gen-z ready to perish Sep 10 '17 at 2:23

Hint. $$\frac{x + 2}{x^2 + x + 1} = \frac{1}{2} \cdot \frac{2x + 1}{x^2 + x + 1} + \frac{1.5}{(x + \frac{1}{2})^2 + \frac{3}{4}}$$

• yes, the next part you should have something with $$arctan \ or \ (tan^-1)$$ – valer Sep 10 '17 at 2:20
• Substitute $$u=x+\frac{1}{2} \ and \ write \frac{3}{4}=(\frac{\sqrt{3}}{2})^2$$ @DevHeavy – valer Sep 10 '17 at 2:31
• yes the last one is to the power of 2 soo you should get the $$\frac{1}{a} \cdot arctan(\frac{u}{a})$$, $$\ where \ in \ our \ case \ a=\frac{\sqrt{3}}{2}$$ @DevHeavy – valer Sep 10 '17 at 2:41
• Yes that is correct @DevHeavy – valer Sep 10 '17 at 2:54
• – Math Lover Sep 10 '17 at 3:12