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Math Help - Once again, another Partial Fractions Problem

  1. #1
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    Once again, another Partial Fractions Problem

    Hey guys, me again. I'm having trouble with another partial fractions problem, except with this one the degree on the top is larger than on the bottom, resulting in division. Here's the problem.

    \int \frac{x^3}{x^2+1}dx
    So what you do is you divide the denominator by the numerator and you get
    x - \frac{x}{x^2+1}
    So now that you have a remainder, you find the integral of that. This is where I'm having trouble, so in otherwords, I'm having trouble with the following problem...

    \int \frac{x}{x^2+1}

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by larson View Post
    Hey guys, me again. I'm having trouble with another partial fractions problem, except with this one the degree on the top is larger than on the bottom, resulting in division. Here's the problem.

    \int \frac{x^3}{x^2+1}dx
    So what you do is you divide the denominator by the numerator and you get
    x - \frac{x}{x^2+1}
    So now that you have a remainder, you find the integral of that. This is where I'm having trouble, so in otherwords, I'm having trouble with the following problem...

    \int \frac{x}{x^2+1}

    Thanks in advance.
    Make the substitution u = x^2 + 1 .......
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    Quote Originally Posted by mr fantastic View Post
    Make the substitution u = x^2 + 1 .......
    Man, I'm really getting to caught up with other things, I'm just not thinking clearly lol. I'll see how the problem goes not with u sub. Thanks though.

    Edit: And I think I got the answer. Thanks again.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by larson View Post
    Man, I'm really getting to caught up with other things, I'm just not thinking clearly lol. I'll see how the problem goes not with u sub. Thanks though.
    incidentally, you could have done mr fantastic's substitution to begin with, and save yourself the polynomial division

    as well as, forget about dividing altogether and use algebraic manipulation. note that \frac {x^3}{x^2 + 1} = \frac {x^3 + x - x}{x^2 + 1} = \frac {x(x^2 + 1) - x}{x^2 + 1} = \frac {x(x^2 + 1)}{x^2 + 1} - \frac x{x^2 + 1} = x - \frac x{x^2 + 1}
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    incidentally, you could have done mr fantastic's substitution to begin with, and save yourself the polynomial division

    as well as, forget about dividing altogether and use algebraic manipulation. note that \frac {x^3}{x^2 + 1} = \frac {x^3 + x - x}{x^2 + 1} = \frac {x(x^2 + 1) - x}{x^2 + 1} = \frac {x(x^2 + 1)}{x^2 + 1} - \frac x{x^2 + 1} = x - \frac x{x^2 + 1}
    Didn't think of that... hmmmm...
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