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Math Help - This Integral is Giving Me Trouble...

  1. #1
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    This Integral is Giving Me Trouble...

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

    I set up a substitution and made u=x^2+1 so I could rewrite the integral as \frac{u-1}{u} but my TI-89 is giving me the answer as x-\arctan{x}

    I was thinking of using a trig-substitution, but this is for a friend of mine's homework, and he hasn't learned those yet. I know it's something simple that I'm missing, but I can't figure it out.
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Kalter Tod View Post
    \int{\frac{x^2}{x^2+1}}

    I set up a substitution and made u=x^2+1 so I could rewrite the integral as \frac{u-1}{u} but my TI-89 is giving me the answer as x-\arctan{x}

    I was thinking of using a trig-substitution, but this is for a friend of mine's homework, and he hasn't learned those yet. I know it's something simple that I'm missing, but I can't figure it out.
    \int \frac{x^2}{x^2 + 1}dx

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

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

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

    = x - arctan(x) + C
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Kalter Tod View Post
    \int{\frac{x^2}{x^2+1}}

    I set up a substitution and made u=x^2+1 so I could rewrite the integral as \frac{u-1}{u} but my TI-89 is giving me the answer as x-\arctan{x}

    I was thinking of using a trig-substitution, but this is for a friend of mine's homework, and he hasn't learned those yet. I know it's something simple that I'm missing, but I can't figure it out.
    Generally when the numerator contains a power that is equal or one degree greater than the denominator you use polynomial division (or this technique that jandvl uses which is what I do) it is the best way to do these
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