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Math Help - Integration by parts, help asap please, exam soon

  1. #1
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    Integration by parts, help asap please, exam soon

    \int x tan^- (x) dx ( The Tan is inverse, i'm not sure how to properly display that in latex )

    i did

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

    Is it right ? If yes

    now how can I solve \int (\frac{x^2}{2}) (\frac{1}{1+x}) dx
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  2. #2
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    The derivative of \tan^{-1}{x} is \frac{1}{1+x^2} and not \frac{1}{1+x}.

    To solve \frac{1}{2}\int \frac{x^2}{1+x^2}~dx, simply add 1/subtract 1 (which is actually a shortcut around a polynomial division) to get:

    \frac{1}{2}\int \frac{\rlap{-------}1+x^2}{\rlap{-------}1+x^2}~dx - \frac{1}{2}\int \frac{1}{1+x^2}~dx
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  3. #3
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    Sorry if I sound a bit ignorant, but I don't really get it, I'm a bit new to all those stuff... could you explain a bit more? What did you add and what did you substract, and then what would i get ?
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  4. #4
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    \int (\frac{x^2}{2}) (\frac{1}{1+x}) dx

    I divided the polynomials in this integral and i got \int 1 - \frac {1}{1+x^2}

    solution for that is x - ln | 1 + x^2 | + c

    is this right ?
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  5. #5
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    the integral of \int \frac{1}{1+x^2}~dx<br />
is \tan^{-1}{x}<br />
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  6. #6
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    so \int 1 - \frac {1}{1+x^2}

    =

    x -


    thanks a lot guys, eventhough i didn't understand Chop Suey method, but I'm glad i made some progress -,- just started learning integrals yesterday

    Cheers
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