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Math Help - Integrate

  1. #1
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    Integrate

    the intergal of 1/(e^(2x)-1) dx
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  2. #2
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    Hello,

    u=e^{2x} and du=2u.dx.

    So \int \frac{1}{1+e^{2x}}dx=\frac{1}{2}\int \frac{1}{u(u-1)}

    But \frac{1}{u(u-1)}=\frac{1}{u-1}-\frac{1}{u}.

    Hence \frac{1}{2}\int \left(\frac{1}{u-1}-\frac{1}{u}\right)=\frac{1}{2}\left(\ln(u-1)-\ln(u)\right)+C

    Therefore a primitive is \color{blue}\boxed{x\mapsto \frac{1}{2}\ln\left(\frac{e^{2x}-1}{e^{2x}}\right)} or still \boxed{x\mapsto \frac{1}{2}\ln(2sh(x))}

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  3. #3
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    Quote Originally Posted by tiga killa View Post
    the intergal of 1/(e^(2x)-1) dx
    \frac{1}{e^{2x}-1} = \frac{1 - e^{2x} + e^{2x}}{e^{2x} - 1} = \frac{e^{2x}}{e^{2x}-1} - \frac{e^{2x}-1}{e^{2x}-1} = \frac{e^{2x}}{e^{2x}-1} - 1

    integrate the last expression
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  4. #4
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    Yes, why not, it's faster.

    Hello skeeter
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