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Math Help - Integration by parts

  1. #1
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    Integration by parts

    \int ln(x^2-4)dx=xln(x^2-4)-\int \frac{x(2x)}{x^2-4}dx

    =xln(x^2-4)-(xln(x^2-4)+\int ln(x^2-4)dx

    =\int ln(x^2-4)dx

    Where did i go wrong?
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  2. #2
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    Re: Integration by parts

    Quote Originally Posted by Punch View Post
    \int ln(x^2-4)dx=xln(x^2-4)-\int \frac{x(2x)}{x^2-4}dx

    =xln(x^2-4)-(xln(x^2-4)+\int ln(x^2-4)dx

    =\int ln(x^2-4)dx

    Where did i go wrong?
    note that \ln(x^2-4) = \ln(x+2) + \ln(x-2)

    I would break the original integrand into the two terms above, then use integration by parts for each integral.
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  3. #3
    Member sbhatnagar's Avatar
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    Re: Integration by parts

    Quote Originally Posted by Punch View Post
    \int ln(x^2-4)dx=xln(x^2-4)-\int \frac{x(2x)}{x^2-4}dx<--------------You did right till here.

    =xln(x^2-4)-(xln(x^2-4)+\int ln(x^2-4)dx

    =\int ln(x^2-4)dx

    Where did i go wrong?
    ...

    You have I=x\ln{(x^2-4)}-\int \frac{2x^2}{x^2-4}dx

    \begin{align*} &=x\ln{(x^2-4)}-2\int \frac{x^2-4+4}{x^2-4}dx = x\ln{(x^2-4)}-2{\left[ x+\int \frac{4}{x^2-4}dx\right]} \end{align*}

    = x\ln{(x^2-4)}-2{\left[ x+\int \frac{1}{x-2}dx-\int\frac{1}{x+2}dx\right]}

    Proceed
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  4. #4
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    Re: Integration by parts

    Quote Originally Posted by sbhatnagar View Post
    ...

    You have I=x\ln{(x^2-4)}-\int \frac{2x^2}{x^2-4}dx

    \begin{align*} &=x\ln{(x^2-4)}-2\int \frac{x^2-4+4}{x^2-4}dx = x\ln{(x^2-4)}-2{\left[ x+\int \frac{4}{x^2-4}dx\right]} \end{align*}

    = x\ln{(x^2-4)}-2{\left[ x+\int \frac{1}{x-2}dx-\int\frac{1}{x+2}dx\right]}

    Proceed
    Thank you, I realised something, integrating an improper fraction is a NO-NO
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