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Math Help - A more difficult integral, general power formula

  1. #1
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    A more difficult integral, general power formula

    Question: Integrate 28-1-16

    (int) 0.8((3 + 2 ln(u))^3) du/u

    My attempt:
    u = (3 + 2 ln u)
    du = 2(1/u) = 2/u

    My guess at this is to divide the whole integral by 2 now since it has been multiplied by 2 in the derivative.

    resulting in this:

    0.4[3 + 2 ln u]/4

    I used a Ti-89 and got a very weird answer. Any clues as to what went wrong? thanks.
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  2. #2
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    Re: A more difficult integral, general power formula

    Hello, togo!

    You are using u for two different quantities.


    \displaystyle\int 0.8(3 + 2\ln u)^3\,\frac{du}{u}

    We have: . 0.8\int(3+2\ln u)^3\,\frac{du}{u}

    Let v \:=\:3+2\ln u \quad\Rightarrow\quad dv \:=\:2\,\frac{du}{u} \quad\Rightarrow\quad \frac{du}{u} \:=\:\tfrac{1}{2}dv

    Substitute: . 0.8\int v^3\left(\tfrac{1}{2}dv\right) \;=\; 0.4\int v^3\,du \;=\;0.1v^4+C

    Back-substitute: . 0.1(3 + 2\ln x)^4 + C
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  3. #3
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    Re: A more difficult integral, general power formula

    thats cool thanks
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