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Math Help - Integration

  1. #1
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    Integration

    Use the substitution x = 1/u to find the value of

     \int_{\frac{1}{2}}^{2} \frac {lnx}{1+x^2}dx


    Please help me to solve this.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    x=\frac{1}{u} \implies \mathrm{d}x=-\frac{\mathrm{d}u}{u^2}

    Substitute this in the integral, it should give something interesting.
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  3. #3
    MHF Contributor red_dog's Avatar
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    \displaystyle x=\frac{1}{u}\Rightarrow dx=-\frac{1}{u^2}du
    \displaystyle x=\frac{1}{2}\Rightarrow t=2, \ x=2\Rightarrow t=\frac{1}{2}.
    So, \displaystyle I=\int_{\frac{1}{2}}^2\frac{\ln x}{1+x^2}dx=\int_2^{\frac{1}{2}}\frac{\ln u}{1+u^2}du=-\int_{\frac{1}{2}}^2\frac{\ln u}{1+u^2}du=-I\Rightarrow I=0
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  4. #4
    MHF Contributor
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    Note how the wording is just a little strange. "to find" isn't the usual task, is it?

    Do the substitution. What happens?

    Note the symmetry of the substitution with the limits of integration.

    Hint: How many Numbers, R, can you think of where R = -R?
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