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Math Help - Indefinite integral

  1. #1
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    Indefinite integral

    I would like some help with this problem:

    Evaluate

    \int \frac{x}{x^3 + 1} dx
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  2. #2
    o_O
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    There's always partial fraction decomposition. Can't think of a shorter way atm .. maybe Krizalid can think of something!

    :

    \int \frac{x}{x^3 + 1} \ dx

     = \int \frac{x}{(x+1)(x^2-x+1)} \ dx

     = \int \left(-\frac{1}{3(x+1)} \ + \ \frac{1}{3} \cdot \frac{x+1}{x^2-x+1}\right) dx (By partial fraction decomposition)

    = \int \left(-\frac{1}{3(x+1)} \ + \ \frac{1}{3} \cdot \frac{x+1 {\color{red} \ + 1 - 1}}{x^2-x+1}\right) dx

    = \int \left(-\frac{1}{3(x+1)} \ + \ \frac{1}{3} \cdot \frac{x-1}{x^2-x+1} \ + \ \frac{2}{3} \frac{1}{\left(x - \frac{1}{2}\right)^2 +  \frac{3}{4}}\right) \ dx

    = \hdots
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  3. #3
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    Quote Originally Posted by o_O View Post
    There's always partial fraction decomposition. Can't think of a shorter way atm .. maybe Krizalid can think of something!

    = \int \left(-\frac{1}{3(x+1)} \ + \ \frac{1}{3} \cdot \frac{x+1 {\color{red} \ + 1 - 1}}{x^2-x+1}\right) dx

    = \int \left(-\frac{1}{3(x+1)} \ + \ \frac{1}{3} \cdot \frac{x-1}{x^2-x+1} \ + \ \frac{2}{3} \frac{1}{\left(x - \frac{1}{2}\right)^2 +  \frac{3}{4}}\right) \ dx

    = \hdots
    I understand the procedure of adding those extra terms but I was wondering if you could explain the calculations that you have done between that step and the next one? Thanks!
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  4. #4
    o_O
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    Angry

    Sure:
    \frac{1}{3}\left(\frac{x+1 {\color{red} \ - 1 + 1}}{x^2-x-1}\right)

    =\frac{1}{3}\left(\frac{x-1}{x^2 - x + 1} + \frac{2}{x^2 - x - 1}\right)

    Completing the square in the bottom of the 2nd fraction:

    =\frac{1}{3}\left(\frac{x-1}{x^2 -x + 1} + \frac{2}{\left(x^2 - x {\color{red}\ + \left(\frac{1}{2}\right)^2} \right) + 1 {\color{red}- \left(\frac{1}{2}\right)^2}}\right)

    = \frac{1}{3} \left(\frac{x-1}{x^2 - x + 1} + \frac{2}{\left(x - \frac{1}{2}\right)^2 + \frac{3}{4}}\right)

    Now multiply the \frac{1}{3} in and you'll get what I had earlier. Use a simple sub for the first one and a trig sub for the other.
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  5. #5
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    Thumbs up

    Thanks, I hope you are not still angry lol
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  6. #6
    o_O
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    Oh don't mind the emoticon. I accidentally pressed it
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  7. #7
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    Quote Originally Posted by o_O View Post

    maybe Krizalid can think of something!
    There's nothing to tell.

    That's the standard solution.
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