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Math Help - Substitution integration help needed

  1. #1
    Member ssadi's Avatar
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    Oct 2008
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    Substitution integration help needed

    Use the substitution x=\frac{1}{u} to find the value of
    \int^2_{\frac{1}{2}} \, \frac{ln x}{1+x^2} dx
    The substitution rounds up to the same expression. Help

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  2. #2
    Member Abu-Khalil's Avatar
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    \frac{1}{z}=x\Rightarrow - \frac{dz}{z^2}=dx then I=\int_\frac{1}{2}^2\frac{\log x}{1+x^2}dx=\int_2^\frac{1}{2}\frac{-\log\frac{1}{z}}{1+\frac{1}{z^2}}\frac{1}{z^2}dz=-\int_\frac{1}{2}^2\frac{\log z}{1+z^2}dz=-I \Rightarrow I=-I\Rightarrow I=0.
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