Integral using partial fractions

1. Oct 29, 2009

manenbu

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

Having difficulties with this one.
I managed to break it down to two partial fractions, being $x-1$ and $x^2+x+1$ but couldn't make anything out of it.

2. Oct 29, 2009

Staff: Mentor

Yes, x3 - 1 = (x - 1)(x2 + x + 1), but what did you get for your partial fractions? They should look like this:
$$\frac{A}{x - 1}~+~\frac{Bx + C}{x^2 + x + 1}$$
You should have/will need to find the values for A, B, and C.

3. Oct 29, 2009

manenbu

Sorry, forgot.

I ended up having this:

$$\frac{1}{3} \int \left(\frac{1-x}{x^2+x+1} + \frac{1}{x-1} \right)dx$$

the right part is easy, it's ln|x-1|, but the left part is what's bothering me. Tried completing the square, separating it into 2 different integrals, nothing helps.

4. Oct 29, 2009

Staff: Mentor

$$\left(\frac{1-x}{x^2+x+1} \right)$$
$$=~(-1/2)~\frac{(2x - 1) - 1}{x^2 + x + 1}$$
Now split this up into two fractions, and integrate each. The first one you can solve by substition, with u = x2 + x + 1.

The second one is a little harder, as you will have to complete the square to get the denominator in the form (x + <something>)2 + <something else>. The idea here is that
$$\int \frac{du}{u^2 + a^2}~=~(1/a)tan^{-1}(u/a) + C$$

5. Oct 30, 2009

manenbu

You had a small mistake here, it should be (2x+1)-3 and not (2x-1)-1.

But thanks! It helped me solve it.