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Math Help - More integral help!

  1. #1
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    More integral help!

    So I've been trying to figure out the best way to approach this problem, but so far no luck.

    (int(x*ln(x))/sqrt(x^2-1)

    I'm thinking this has something to do with inverse sin, but I am not sure what to substitute...

    Help please?
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  2. #2
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    if u = \sqrt{x^2-1} then du = \dfrac{x\, dx}{\sqrt{x^2-1}} giving

    \displaystyle \dfrac{1}{2}\int \ln \left(u^2 + 1\right)\, du then try parts.
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  3. #3
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    Quote Originally Posted by shaunthepanda View Post
    So I've been trying to figure out the best way to approach this problem, but so far no luck.

    (int(x*ln(x))/sqrt(x^2-1)

    I'm thinking this has something to do with inverse sin, but I am not sure what to substitute...

    Help please?
    Integration by parts

    u = ln(x)

    dv = \frac{1}{\sqrt{x^2-1}}
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  4. #4
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    Ok so if u=ln(x) then dv should equal x/sqrt(x^2-1) but if that is the case then how do I integrate x/sqrt(x^2-1)?
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  5. #5
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    Quote Originally Posted by shaunthepanda View Post
    Ok so if u=ln(x) then dv should equal x/sqrt(x^2-1) but if that is the case then how do I integrate x/sqrt(x^2-1)?
    \displaystyle \int \frac{x}{\sqrt{x^2-1}} \, dx = \sqrt{x^2-1} + C
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  6. #6
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    Quote Originally Posted by shaunthepanda View Post
    Ok so if u=ln(x) then dv should equal x/sqrt(x^2-1) but if that is the case then how do I integrate x/sqrt(x^2-1)?
    Here is what I was thinking.

    \displaystyle u=\ln{x} \ \ \ \ dv=\frac{1}{\sqrt{x^2-1}}

    \displaystyle du=\frac{dx}{x} \ \ \ \ v=\ln{(\sqrt{x^2-1}+x)}

    \displaystyle uv-\int vdu=\ln{(x)}\cdot\ln{(\sqrt{x^2-1}+x)}-\int\frac{\ln{(\sqrt{x^2-1}+x)}}{x}dx
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  7. #7
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    Quote Originally Posted by dwsmith View Post
    Here is what I was thinking.

    \displaystyle u=\ln{x} \ \ \ \ dv=\frac{1}{\sqrt{x^2-1}}

    \displaystyle du=\frac{dx}{x} \ \ \ \ v=\ln{(\sqrt{x^2-1}+x)}

    \displaystyle uv-\int vdu=\ln{(x)}\cdot\ln{(\sqrt{x^2-1}+x)}-\int\frac{\ln{(\sqrt{x^2-1}+x)}}{x}dx
    You seem to have neglected the first "x" in the integral:
    \int x\frac{ln(x}}{\sqrt{x^2- 1}}dx
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  8. #8
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    Shouldn't it be \displaystyle u = \ln{x} and \displaystyle dv = \frac{x}{\sqrt{x^2 - 1}}\,dx?
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  9. #9
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    Quote Originally Posted by Prove It View Post
    Shouldn't it be \displaystyle u = \ln{x} and \displaystyle dv = \frac{x}{\sqrt{x^2 - 1}}\,dx?
    Quote Originally Posted by HallsofIvy View Post
    You seem to have neglected the first "x" in the integral:
    \int x\frac{ln(x}}{\sqrt{x^2- 1}}dx
    Try my integration and with the x. They should yield the same results.
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