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Math Help - integral (ln(x))^2

  1. #1
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    integral (ln(x))^2

    Man I'm having problems.
    Question is
     \int_1^2 {(\ln (x))^2 } dx \\

    I get this
    = \int_1^2 {\ln (x)\ln (x)dx}  \\
     = \frac{1}{x}\ln (x) - \int {\frac{1}{x}}  \times \frac{1}{x}dx \\
      = \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\
      = \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\
    = \frac{1}{x}\ln (x) + \frac{1}{x} \\
      = [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\
    = \frac{1}{2}\ln (2) - \frac{1}{2} \\<br />
    But apparently it should be
     = 2(\ln (2))^2  - 4\ln (2) + 2<br />
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  2. #2
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    Quote Originally Posted by Craka View Post
    Man I'm having problems.
    Question is
     \int_1^2 {(\ln (x))^2 } dx \\

    I get this
    = \int_1^2 {\ln (x)\ln (x)dx} \\
     = \frac{1}{x}\ln (x) - \int {\frac{1}{x}} \times \frac{1}{x}dx \\
     = \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\
     = \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\
    = \frac{1}{x}\ln (x) + \frac{1}{x} \\
     = [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\
    = \frac{1}{2}\ln (2) - \frac{1}{2} \\<br />
    But apparently it should be
     = 2(\ln (2))^2 - 4\ln (2) + 2<br />
    You have not applied the integration by parts forumla correctly. Go back and check what it says.

    There are many ways of doing it. The best way for you I think is to make the substitution u = \ln x. Then the integral becomes \int_{0}^{\ln 2} u^2 \, e^u \, du. Now use integration by parts - twice.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Craka View Post
    Man I'm having problems.
    Question is
     \int_1^2 {(\ln (x))^2 } dx \\

    I get this
    = \int_1^2 {\ln (x)\ln (x)dx}  \\
     = \frac{1}{x}\ln (x) - \int {\frac{1}{x}}  \times \frac{1}{x}dx \\
      = \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\
      = \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\
    = \frac{1}{x}\ln (x) + \frac{1}{x} \\
      = [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\
    = \frac{1}{2}\ln (2) - \frac{1}{2} \\<br />
    But apparently it should be
     = 2(\ln (2))^2  - 4\ln (2) + 2<br />
    See here

    personally, i like my way the best (this is usually not the case)

    ...oh, i didn't show the full working... oh well, i gave a small outline. try it. Mr F's method is there too

    There should be another thread around where this was done completely...you can do a search if you want
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Craka View Post
    Man I'm having problems.
    Question is
     \int_1^2 {(\ln (x))^2 } dx \\

    I get this
    = \int_1^2 {\ln (x)\ln (x)dx} \\
     = \frac{1}{x}\ln (x) - \int {\frac{1}{x}} \times \frac{1}{x}dx \\
     = \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\
     = \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\
    = \frac{1}{x}\ln (x) + \frac{1}{x} \\
     = [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\
    = \frac{1}{2}\ln (2) - \frac{1}{2} \\<br />
    But apparently it should be
     = 2(\ln (2))^2 - 4\ln (2) + 2<br />
    \int\ln^2(x)dx

    Let \ln(x)=u\Rightarrow{x=e^u}

    So dx=e^u

    Giving

    \int{e^uu^2du}

    Parts gives

    \int{e^uu^2du}=u^2e^u-2\int{e^u\cdot{u}du}=u^2e^u-2\bigg[e^uu-\int{e^u}\bigg]=u^2e^u-2ue^u+2e^u= (u^2-2u+2)e^u

    Now back subbing we get

    \int\ln^2(x)=(\ln^2(x)-2\ln(x)+2)x+C

    EDIT: Sorry Mr. F! Didn't even see that you had that there! Well, at least the poster sees what you meant
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Craka View Post
    Man I'm having problems.
    Question is
     \int_1^2 {(\ln (x))^2 } dx \\

    I get this
    = \int_1^2 {\ln (x)\ln (x)dx} \\
     = \frac{1}{x}\ln (x) - \int {\frac{1}{x}} \times \frac{1}{x}dx \\
     = \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\
     = \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\
    = \frac{1}{x}\ln (x) + \frac{1}{x} \\
     = [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\
    = \frac{1}{2}\ln (2) - \frac{1}{2} \\<br />
    But apparently it should be
     = 2(\ln (2))^2 - 4\ln (2) + 2<br />
    But if you were to do it your way

    let u=\ln(x)
    and math dv=\ln(x)dx
    So du=\frac{dx}{x}
    and v=\int\ln(x)dx=x\ln(x)-x=x(\ln(x)-1)

    Giving us

    \int\ln^2(x)=\ln(x)x(\ln(x)-1)-\int\frac{x(\ln(x)-1)}{x}dx

    So for the second integral we get

    \int\frac{x(\ln(x)-1)}{x}dx=\int\ln(x)-1dx=x(\ln(x)-2)

    Giving us a grand finale of

    \int\ln^2(x)=x\ln(x)(\ln(x)-1)-x(\ln(x)-2)=x(\ln^2(x)-2\ln(x)+2)+C

    An easier method that might have been mentioned elsewhere would be as if you were just calculating \int\ln(x)dx

    So for this case let

    u=\ln^2(x)
    and dv=dx
    so then
    du=\frac{2\ln(x)}{x}dx
    and v=x

    So once again using our parts formula we get

    \int\ln^2(x)dx=\ln^2(x)\cdot{x}-\int\frac{2\ln(x)\cdot{x}}{x}

    Now for the second one we apply parts again and once again let dv=dx to get

    \int\ln^2(x)dx=\ln^2(x)\cdot{x}-2(\ln(x)\cdot{x}-x)=(\ln^2(x)-2\ln(x)+2)x+C

    as you can see your method is by far the messiest, there are very few integrals where you have \int{u^2(x)dx}

    and let u=u(x) and dv=u(x)

    There are some I think, but they are uncommon
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