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Math Help - integral

  1. #1
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    integral

    question in attachment
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    None of them.
    E?

    RonL
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  3. #3
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    Hello, DINOCALC09!

    Did you do the substitution?
    (I agree with the Captain.)


    If the substitution u = \sqrt{x-1} is made, the integral . \int^5_2\frac{\sqrt{x-1}}{x}\,dx\;=\

    We have: . u \:=\: \sqrt{x-1}\quad\Rightarrow\quad x \:=\:u^2-1\quad\Rightarrow\quad dx \:=\;2u\,du

    Substitute: . \int\frac{u}{u^2+1}(2u\,du) \;=\;\int\frac{2u^2}{u^2+1}\,du

    Change limits: . \begin{Bmatrix}x = 2 & \Rightarrow & u = 1 \\ x = 5 & \Rightarrow & u = 2\end{Bmatrix}

    . . Answer: . \int^2_1\frac{2u^2}{u+1}\,du . (E)

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