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Math Help - Integration

  1. #1
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    Integration

    Find ∫ (ln x) ^2 dx.

    Thank you!!!
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  2. #2
    o_O
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    Use integration by parts:

    \begin{array}{ll}u = \big[\ln(x)\big]^{2} & dv = dx \\ du = 2\ln(x) \cdot \frac{1}{x} & v = x\end{array}

    So,
    \int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - \int \left(2\ln(x) \cdot \frac{1}{x} \cdot x\right) dx
    \int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - 2{\color{blue}\int \left( \ln(x)\right) dx}

    The thing in blue needs integration by parts again but I'll leave it up to you with that one (use a similar sub as we did earlier!).
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Tangera View Post
    Find ∫ (ln x) ^2 dx.

    Thank you!!!
    This was my first thought...and I'm gonna go along with it...

    \int (\ln(x))^2\,dx

    let u=\ln(x)\implies x=e^u

    thus, du=\frac{\,dx}{x}\implies \,dx=x\,du

    Thus, \int(\ln(x))^2\,dx=\int u^2e^{u}\,du

    Apply integration by parts...

    w=u^2 and \,dv=e^{u}\,du

    \therefore \,dw=2u\,du and v=e^{u}

    Thus, we get -u^2e^{u}-2\int ue^{u}\,du

    let w=u and \,dv=e^{u}\,du

    \therefore \,dw=\,du and v=e^{u}

    Now we see that we get

    u^2e^{u}-2\left[ue^{u}-\int e^{u}\,du\right]=e^{u}\left(u^2-2u+2\right)

    Substitute u=\ln(x) back in to get

    \int (\ln(x))^2\,dx=\color{red}\boxed{x\left[(\ln(x))^2-2\ln(x)+2\right]}

    Does this make sense?

    --Chris

    EDIT: o_O beat me to it!
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by o_O View Post
    Use integration by parts:

    \begin{array}{ll}u = \big[\ln(x)\big]^{2} & dv = dx \\ du = 2\ln(x) \cdot \frac{1}{x} & v = x\end{array}

    So,
    \int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - \int \left(2\ln(x) \cdot \frac{1}{x} \cdot x\right) dx
    \int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - 2{\color{blue}\int \left( \ln(x)\right) dx}

    The thing in blue needs integration by parts again but I'll leave it up to you with that one (use a similar sub as we did earlier!).
    Why do I have to make life so hard?

    --Chris
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