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Math Help - One quick integral

  1. #1
    MHF Contributor Mathstud28's Avatar
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    One quick integral

    I have seen this integral done in so many ways, and I am curious as to how each person (if anyone even does it) computes it on this site. I think it gives insight into that persons mathematical prefereneces. The integral is


    \int\sqrt{a+be^{cx}}dx

    Where a,b,c are constants
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    I'd do z^2=a+be^{cx}.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Yeah me too, Here is my solution if anyone is interested

    \int\sqrt{a+be^{cx}}dx

    Let \psi=\sqrt{a+be{cx}}\Rightarrow\frac{1}{c}\ln\left  (\frac{\psi^2-a}{b}\right)=x

    So

    dx=\frac{2}{c}\frac{\psi}{\psi^2-a}d\psi

    So our integral turns into

    \frac{2}{c}\int\frac{\psi^2}{\psi^2-a}d\psi

    Now adding 0 we get

    \frac{2}{c}\bigg[\int\frac{\psi^2-a}{\psi^2-a}d\psi-a\int\frac{d\psi}{a-\psi^2}\bigg]

    Now for the second integral I rewrote this as

    a\frac{1}{a}\int\frac{d\psi}{1-\left(\frac{\psi}{\sqrt{a}}\right)^2}

    I then let \varphi=\frac{\psi}{\sqrt{a}}\Rightarrow{d\psi=\sq  rt{a}\varphi}

    So we have

    \sqrt{a}\int\frac{d\varphi}{1-\varphi^2}=\sqrt{a}{\rm{arctanh}}(\varphi)

    Back subbing we get

    \sqrt{a}{\rm{arctanh}}\left(\frac{\psi}{\sqrt{a}}\  right)

    So we have

    \frac{2}{c}\bigg[\psi-\sqrt{a}{\rm{arctanh}}\left(\frac{\psi}{\sqrt{a}}\  right)\bigg]

    Now for our final backsub we get

    \boxed{\int\sqrt{a+be^{cx}}dx=\frac{2}{c}\bigg[\sqrt{a+be^{cx}}-\sqrt{a}{\rm{arctanh}}\left(\frac{\sqrt{a+be^{cx}}  }{\sqrt{a}}\right)\bigg]+C}
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