integration by partial fractions problem

integral of [1 / (1 + y^3)] dy

1 / (1 + y^3) = A / (y + 1) + (Bx + C) / (y^2 - y + 1)

so i found A = 1/3, B = -1/3, and C = 2/3 assuming that i found the correct values.

so the integral now looks like (1/3) / (y+1) + (-x/3 + 2/3) / (y^2 - y + 1)

i integrated the first part (1/3) / (y+1) and got 1/3 ln|y + 1|. but i can't seem to integrate the second part (-x/3 + 2/3) / (y^2 - y + 1).

i let u = y^2 - y + 1 and i found that du = 2y - 1 dy. however i can't make the top look like 2y - 1 and get rid of the dy. please help!