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Math Help - integration problem

  1. #1
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    Question integration problem

    Could use some help here:

    1. The integral: 1 / sqrt( e^x - 1 ) dx

    2. The integral: sqrt( x^2 - 1) / x dx

    Thanks!
    Last edited by coolio; February 25th 2008 at 01:13 PM.
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  2. #2
    Super Member PaulRS's Avatar
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    We have \int{\frac{dx}{e^x-1}}=\int{\frac{e^x}{(e^x-1)\cdot{e^x}}dx}

    Let u=e^x then \frac{du}{dx}=e^x

    We get: \int{\frac{dx}{e^x-1}}=\int{\frac{du}{(u-1)\cdot{u}}}

    See that \frac{1}{(u-1)\cdot{u}}=\frac{1}{u-1}-\frac{1}{u}

    So: \int{\frac{du}{(u-1)\cdot{u}}}=\int{\frac{du}{u-1}}-\int{\frac{du}{u}}=\ln(1-1/u)+k

    \int{\frac{dx}{e^x-1}}=\ln\left(1-\frac{1}{e^x}\right)+k
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  3. #3
    Senior Member Peritus's Avatar
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    I think you missed the sqrt


    \begin{gathered}<br />
  \int {\frac{1}<br />
{{\sqrt {e^x  - 1} }}dx}  \hfill \\<br />
  u^2  = e^x  - 1 \hfill \\<br />
  2udu = e^x dx \hfill \\ <br />
\end{gathered}

    <br />
\int {\frac{1}<br />
{u}\frac{{2u}}<br />
{{u^2  + 1}}du}  = 2\int {\frac{1}<br />
{{1 + u^2 }}du = 2\arctan u + C = 2\arctan \sqrt {e^x  - 1}  + C} <br />


    -------------------------------------------------------------------------

    <br />
\begin{gathered}<br />
  \int {\frac{{\sqrt {x^2  - 1} }}<br />
{x}dx}  \hfill \\<br />
  x^2  = u^2  + 1 \hfill \\<br />
  xdx = udu \hfill \\ <br />
\end{gathered} <br />

    <br />
\int {\frac{{u^2 }}<br />
{{u^2  + 1}}du = } \int {1 - \frac{1}<br />
{{u^2  + 1}}du = u - \arctan u + C = \sqrt {x^2  - 1}  - \arctan \sqrt {x^2  - 1}  + C} <br />
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  4. #4
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    Krizalid's Avatar
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    Anyway if the problem would've been

    Quote Originally Posted by PaulRS View Post
    We have \int{\frac{dx}{e^x-1}}
    then

    \int {\frac{1}<br />
{{e^x  - 1}}\,dx}  = \int {\frac{{e^x  - \left( {e^x  - 1} \right)}}<br />
{{e^x  - 1}}\,dx}  = \int {\frac{{\left( {e^x  - 1} \right)'}}<br />
{{e^x  - 1}}\,dx}  - \int {dx}.

    Hence \int {\frac{1}<br />
{{e^x  - 1}}\,dx}  = \ln \left| {e^x  - 1} \right| - x + k.
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  5. #5
    Senior Member Peritus's Avatar
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    Quote Originally Posted by Krizalid View Post
    Anyway if the problem would've been


    then

    \int {\frac{1}<br />
{{e^x  - 1}}\,dx}  = \int {\frac{{e^x  - \left( {e^x  - 1} \right)}}<br />
{{e^x  - 1}}\,dx}  = \int {\frac{{\left( {e^x  - 1} \right)'}}<br />
{{e^x  - 1}}\,dx}  - \int {dx}.

    Hence \int {\frac{1}<br />
{{e^x  - 1}}\,dx}  = \ln \left| {e^x  - 1} \right| - x + k.
    Very nice, made me smile.
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  6. #6
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    Thanks a for the help everyone!

    I came across another one that has stumped me:

    The integral from -1 to 1: sinx / 1 + x^2 dx

    I see 1 / 1+x^2 as the derivative of arctan, but I'm not sure if that is relevant. The only this I could think of was to integrate by parts, but after attempting that, I seem to just be going in circles.
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  7. #7
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    It's a classic problem, the answer is zero, since you're integrating an odd function on a symmetric interval.
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  8. #8
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    Quote Originally Posted by Krizalid View Post
    It's a classic problem, the answer is zero, since you're integrating an odd function on a symmetric interval.
    Thank you for the reply but could you expand on that a bit? I'm not entirely sure what you mean by odd function.
    Last edited by coolio; February 25th 2008 at 03:00 PM. Reason: typo
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  9. #9
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    Let k=\int_{ - 1}^1 {\frac{{\sin x}}<br />
{{1 + x^2 }}\,dx}, substitute u=-x,

    k =  - \int_{ - 1}^1 {\frac{{\sin u}}<br />
{{1 + u^2 }}\,du} .

    Hence, our original integral satisfies k+k=0\implies k=0.

    (Ahh, I just saw you edited your original question: an odd function f(x) satisfies f(-x)=-f(x). When having integrals whose function is odd besides if they're taken on a symmetric interval [-a,a] the answer is zero.)
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  10. #10
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    Quote Originally Posted by Krizalid View Post
    Let k=\int_{ - 1}^1 {\frac{{\sin x}}<br />
{{1 + x^2 }}\,dx}, substitute u=-x,

    k =  - \int_{ - 1}^1 {\frac{{\sin u}}<br />
{{1 + u^2 }}\,du} .

    Hence, our original integral satisfies k+k=0\implies k=0.
    Ok thanks a lot. That one kind of caught me off guard, as I dont remember seeing anything like that in earlier chapter assignments. I've just been doing the chapter review problems in my text.
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  11. #11
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    Quote Originally Posted by Krizalid View Post
    Let k=\int_{ - 1}^1 {\frac{{\sin x}}<br />
{{1 + x^2 }}\,dx}, substitute u=-x,

    k =  - \int_{ - 1}^1 {\frac{{\sin u}}<br />
{{1 + u^2 }}\,du} .

    Hence, our original integral satisfies k+k=0\implies k=0.
    Ok thanks a lot. That one kind of caught me off guard, as I dont remember seeing anything like that in earlier chapter assignments. It was in the chapter review problems in my text.
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