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Math Help - Integral [1 to 2] (lnx)^2/(x^3) Please help

  1. #1
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    Integral [1 to 2] (lnx)^2/(x^3) Please help

    Need help solving integral [1 to 2] (lnx)^2/(x^3) Please im really stuck
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  2. #2
    Behold, the power of SARDINES!
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    let u=\ln(x) \iff e^{u}=x \iff e^{2u}=x^2 \iff e^{-2u}=\frac{1}{x^2}
    and
    du=\frac{1}{x}dx this gives

    \displaystyle \int_{1}^{2}(\ln(x))^2dx=\int_{0}^{\ln(2)}u^2e^{-2u}du

    Now just integrate by parts twice.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    let u=\ln(x) \iff e^{u}=x \iff e^{2u}=x^2 \iff e^{-2u}=\frac{1}{x^2}
    and
    du=\frac{1}{x}dx this gives

    \displaystyle \int_{1}^{2}(\ln(x))^2dx=\int_{0}^{\ln(2)}u^2e^{-2u}du

    Now just integrate by parts twice.
    So:
    ∫ x^2 * e^(2x) dx = uv - ∫ v du
    = x^2*e^(2x)/2 - ∫ e^(2x)/2 (2x dx)
    = x^2*e^(2x)/2 - ∫ x*e^(2x) dx

    Letting:

    u = x, du = dx
    dv = e^(2x), v = e^(2x)/2 dx



    ∫ x^2 * e^(2x) dx = x^2*e^(2x)/2 - (uv - ∫ v du)
    = x^2*e^(2x)/2 - [x*e^(2x)/2 - ∫ e^(2x)/2 dx]
    = x^2*e^(2x)/2 - x*e^(2x)/2 + 1/2 ∫ e^(2x) dx
    = x^2*e^(2x)/2 - x*e^(2x)/2 + e^(2x)/4 + C
    = e^(2x)/4 * (2x^2 - 2x + 1) + C ?? Is this correct
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by TheEmptySet View Post
    let u=\ln(x) \iff e^{u}=x \iff e^{2u}=x^2 \iff e^{-2u}=\frac{1}{x^2}
    and
    du=\frac{1}{x}dx this gives

    \displaystyle \int_{1}^{2}(\ln(x))^2dx=\int_{0}^{\ln(2)}u^2e^{-2u}du

    Now just integrate by parts twice.
    Or you could just do parts twice without any substitution.
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  5. #5
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    Have a look at this website.
    Be sure to click "show steps".
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by kensington View Post
    Need help solving integral [1 to 2] (lnx)^2/(x^3) Please im really stuck
    If the function to be integrated is...

    \displaystyle f(x)= \frac{\ln^{2} x}{x^{3}} (1)

    ... then integrating by parts three times You obtain...

    \displaystyle \int \frac{\ln^{2} x}{x^{3}}\ dx = - \frac{\ln^{2} x}{2 x^{2}} - \frac{\ln x}{2 x^{2}} - \frac{1}{4 x^{2}} + c  (2)

    ... so that is...

    \displaystyle \int_{1}^{2} \frac{\ln^{2} x}{x^{3}}\ dx = - \frac {\ln^{2} 2 + \ln 2}{8} - \frac{1}{16} + \frac{1}{4} =  .040799756902... (3)

    Kind regards

    \chi \sigma
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