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Math Help - 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:

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


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

    Integration by substitution...

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

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

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

    An online integration calculator gave me:

    -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:

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


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

    Integration by substitution...

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

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

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

    An online integration calculator gave me:

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

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

    \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

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

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

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