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Math Help - more integrals

  1. #1
    Junior Member scottie.mcdonald's Avatar
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    more integrals

    still having problems with integrals. i'm stumped on how to get the four substitutions. here's the problem

    evaluate the integral.

    <br />
\int_1^2 \frac{(lnx)^2dx}{x^3}<br />

    just getting  f\prime(x), f(x), g\prime(x) and g(x) would be a great help. thank you
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by scottie.mcdonald View Post
    still having problems with integrals. i'm stumped on how to get the four substitutions. here's the problem

    evaluate the integral.

    <br />
\int_1^2 \frac{(lnx)^2dx}{x^3}<br />

    just getting  f\prime(x), f(x), g\prime(x) and g(x) would be a great help. thank you
    u=[\ln(x)]^2 \implies du=2\frac{\ln(x)}{x}dx

    dv=\frac{1}{x^3} \implies v=-\frac{1}{2x^2}

    \int_1^2 \frac{(lnx)^2dx}{x^3}=-\frac{[\ln(x)]^2}{2x^2}\bigg|_{1}^{2}+\int_{1}^{2}\frac{\ln(x)}{  x^3}dx

    To finish this you need to integrate by parts again.

    Good luck.
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  3. #3
    Junior Member scottie.mcdonald's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    u=[\ln(x)]^2 \implies du=2\frac{\ln(x)}{x}dx

    dv=\frac{1}{x^3} \implies v=-\frac{1}{2x^2}

    \int_1^2 \frac{(lnx)^2dx}{x^3}=-\frac{[\ln(x)]^2}{2x^2}\bigg|_{1}^{2}+\int_{1}^{2}\frac{\ln(x)}{  x^3}dx
    the  f\prime(x), f(x), g\prime(x) and g(x) seem correct but when you put them in the formula

     \int_1^2 g\prime(x)f(x) = f(x)g(x)|_1^2 - \int_1^2 f\prime(x)g(x) , i am getting something different. the last integral of what I think is different then what you put down ie:

     \int_1^2 2(lnx)(\frac{1}{x})(\frac{1}{2x^3})dx <br />
= \int_1^2 \frac{2(lnx)dx}{2x^4} <br />

    are you calculating a different way?
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