How do I integrate $\frac{x}{1+x^3}$?

Question

I need to evaluate the following:

$$\int_0^1 \dfrac{x}{1+x^3}dx$$

I've managed to get it down to this:

$$-\int_0^1 \dfrac{1}{x+1}dx + \int_0^1 \dfrac{x-1}{x^2-x+1}dx$$

but now I can't integrate the 2nd integral :(

Any hints/tips?

Thanks

Answer 1

You can further separate the second integral as follows: $$\int_0^1 \frac{x-1}{x^2-x+1}\,dx=\frac{1}{2}\left(\int_0^1 \frac{2x-1}{x^2-x+1}\,dx-\int_0^1 \frac{1}{x^2-x+1}\,dx\right)$$ The first integral can be handled using the substitution $x^2-x+1=u$ and the second one can be done by completing the square in denominator.