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Math Help - integration by partial fractions problem

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    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!
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  2. #2
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    Quote Originally Posted by oblixps View Post
    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.
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