How do I evaluate the integral $\int \frac{1}{x^3 -1}dx$

Question

this is my first question here so I hope I did everything right. Still really new to LaTeX as well$$\int \frac{1}{x^3 -1}dx $$

I first used partial fractions to decompose this integral into two parts$$\int \frac{1}{3(x-1)}dx -\frac{1}{3} \int \frac{x+2}{x^2 + x + 1}dx$$

I definitely can solve the first part as $\frac{\ln(x-1)}{3}$ but I'm stuck with the second part as I don't know what to do since the denominator is not easily factor-able.

Thanks in advance to anybody taking the time to read this.

Edit: I do know that you can complete the square to make the integral prettier, but I don't know how to cancel the numerator or seperate it to relate the integral to $\int \frac{1}{x^2 + 1}$ to get some value of arctan(x) +c

Answer 1

Continuing from that part, you break up the integral into two parts:
$$\int \frac u {u^2 + 3/4}du+\int \frac {\frac 3 4} {u^2+\frac 3 4}$$

The first one you do as a simple $u$-substiution to get to $\frac 1 2 \ln (u^2 +\frac 3 4)$.

The second we factor out a $\frac 3 4$ and put the factored out part with the square, to get $$\frac 3 4 \int \frac 1 {(\frac {\sqrt 3} 2u)^2+1}du.$$ From there, use another u substution, setting $v=\frac {\sqrt 3} 2u$ so $du=\frac 2 {\sqrt 3}dv$, getting $$\frac 3 4 \cdot \frac 2 {\sqrt 3}\int \frac 1 {v^2+1}dv,$$ then you get your arctan.

Answer 2

$$ x^2 + x + 1 = \left(x+\frac 1 2\right)^2 + \frac 3 4 = u^2 + \frac 3 4. $$ The standard technique in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial with no first-degree term to is completing the square.