I'm just start with the subject of partial-fractions and I get stuck in this excercise.

Question

I have:

$$\int \frac{1}{x^3-1} dx$$

So...I reach this point, and then I just don´t know how to continue:

$$\frac{1}{3}\ln|x-1|- \frac{1}{3}\int \frac{x+2}{x^2+x+1} dx$$

The answer should be easy, but I don´t know how to get it. Please step by step.

Answer 1

Hint: Since $x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$ we have \begin{align} \int\frac{x+2}{x^2+x+1}\,dx&=\frac{1}{2}\int\frac{2x+4}{x^2+x+1}\,dx\\ &=\frac{1}{2}\int\frac{2x+1+3}{x^2+x+1}\,dx\\ &=\frac{1}{2}\int\frac{2x+1}{x^2+x+1}\,dx+\frac{1}{2}\int\frac{3}{x^2+x+1}\,dx\\ &=\frac{1}{2}\ln\left(x^2+x+1\right)+\frac{3}{2}\int\frac{dx}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\\ &=\frac{1}{2}\ln\left(x^2+x+1\right)+\frac{3}{2}\frac{2}{\sqrt{3}}\arctan\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c\\ &=\frac{1}{2}\ln\left(x^2+x+1\right)+\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)+c \end{align}

Answer 2

Hint: $$\int\frac{\mathrm d\mkern1mu x}{x^2+a^2}=\frac1a\arctan\frac xa.$$