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Thread: Integration by parts, help asap please, exam soon

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

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

    i did

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

    Is it right ? If yes

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

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

    $\displaystyle \frac{1}{2}\int \frac{\rlap{-------}1+x^2}{\rlap{-------}1+x^2}~dx - \frac{1}{2}\int \frac{1}{1+x^2}~dx$
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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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    $\displaystyle \int (\frac{x^2}{2}) (\frac{1}{1+x}) dx$

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

    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 $\displaystyle \int \frac{1}{1+x^2}~dx
    $ is $\displaystyle \tan^{-1}{x}
    $
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  6. #6
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    so $\displaystyle \int 1 - \frac {1}{1+x^2}$

    =

    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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