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Thread: help with this antiderivative

  1. #1
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    help with this antiderivative

    Where did I go wrong here...

    I need to find this AD:

    $\displaystyle \int \frac{1-ix}{1+ix}dx$


    $\displaystyle \int \frac{1-ix}{1+ix}dx= \int \frac{i+x}{i-x}dx$

    Integration by substitution...

    $\displaystyle \int \frac{i+x}{i-x}dx= (i+x)\ln(i-x)-\int \ln(i-x)dx$

    $\displaystyle =(i+x)\ln(i-x)-(i-x)\ln(i-x)-i+x$

    $\displaystyle =2x\ln(i-x)-i+x$

    An online integration calculator gave me:

    $\displaystyle -2i\ln(x-i)-x$

    I can't figure out where my mistake is.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by rainer View Post
    Where did I go wrong here...

    I need to find this AD:

    $\displaystyle \int \frac{1-ix}{1+ix}dx$


    $\displaystyle \int \frac{1-ix}{1+ix}dx= \int \frac{i+x}{i-x}dx$

    Integration by substitution...

    $\displaystyle \int \frac{i+x}{i-x}dx= (i+x)\ln(i-x)-\int \ln(i-x)dx$

    $\displaystyle =(i+x)\ln(i-x)-(i-x)\ln(i-x)-i+x$

    $\displaystyle =2x\ln(i-x)-i+x$

    An online integration calculator gave me:

    $\displaystyle -2i\ln(x-i)-x$

    I can't figure out where my mistake is.
    your mistake in

    $\displaystyle \int \frac{i+x}{i-x}dx= -(i+x)\ln(i-x)+ \int \ln(i-x)dx$

    how you make the next step

    you can solve it like this

    $\displaystyle \int \frac{i+x}{i-x} dx = - \int \frac{i+x}{x-i} dx $

    $\displaystyle -\int \frac{x-i + 2i }{x-i } dx = - \int 1 + \frac{2i}{x-i} dx $

    $\displaystyle -\int 1 dx - \int \frac{2i}{x-i} dx = -x - 2i\ln (x-1) $
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