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Math Help - integration by substitution

  1. #1
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    integration by substitution

    \int\frac{1}{e^x+e^{-x}}dx { let u=e^x}

    \int\frac{1}{e^x+e^{-x}}dx=\int\frac{1}{u+\frac{1}{u}}\frac{1}{u}du

    =\int{1}{u^2+1}du

    =\frac{1}{2}\int\frac{2}{u^2+1}du

    =\frac{1}{2}ln|u^2+1|+c

    =\frac{1}{2}ln|e^{2x}+1|+c
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  2. #2
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    Re: integration by substitution

    It's a good idea to get a common denominator...

    \displaystyle \begin{align*}\int{\frac{1}{e^x + e^{-x}}\,dx} &= \int{\frac{1}{e^x + \frac{1}{e^x}}\,dx} \\ &= \int{\frac{1}{\frac{e^{2x} + 1}{e^x}}\,dx} \\ &= \int{\frac{e^x}{e^{2x} + 1}\,dx} \end{align*}

    Now make the substitution \displaystyle u = e^x \implies du = e^x\,dx and the integral becomes

    \displaystyle \begin{align*} \int{\frac{1}{u^2 + 1}\,du} &= \arctan{u} + C  \\ &= \arctan{\left(e^x\right)} + C\end{align*}
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  3. #3
    Super Member Quacky's Avatar
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    Re: integration by substitution

    Quote Originally Posted by Punch View Post
    \int\frac{1}{e^x+e^{-x}}dx { let u=e^x}

    \int\frac{1}{e^x+e^{-x}}dx=\int\frac{1}{u+\frac{1}{u}}\frac{1}{u}du

    =\int{1}{u^2+1}du

    =\frac{1}{2}\int\frac{2}{u^2+1}du

    Here is your error. The integration here is not correct. You would need something in terms of 'u' on the numerator to be able to do something along these lines.

    =\frac{1}{2}ln|u^2+1|+c

    =\frac{1}{2}ln|e^{2x}+1|+c
    Editted.
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