integration by partial fractions problem

• Jun 19th 2009, 06:08 PM
oblixps
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!
• Jun 19th 2009, 06:20 PM
mr fantastic
Quote:

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!

Note the correction of typos in red. The second part can be written $- \frac{1}{3} \left( \frac{y - 2}{y^2 - y + 1}\right)$.

Now note that $\frac{y - 2}{y^2 - y + 1} = \frac{1}{2} \left( \frac{2y - 4}{y^2 - y + 1} \right) = \frac{1}{2} \left( \frac{2y - 1 - 3}{y^2 - y + 1} \right)$

$= \frac{1}{2} \left( \frac{2y - 1}{y^2 - y + 1} - \frac{3}{y^2 - y + 1}\right)$.

Now note that $\frac{3}{y^2 - y + 1} = \frac{3}{\left(y - \frac{1}{2}\right)^2 + \frac{3}{4}}$.

Your job is to put all these bits and pieces together.