Originally Posted by
oblixps integral of [1 / (1 + y^3)] dy
1 / (1 + y^3) = A / (y + 1) + (B y + 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) + (-y}/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 (-y/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!