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Math Help - U substitution indefinite integral

  1. #1
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    U substitution indefinite integral

    I have been trying to solve the indefinite integral (x-3)*sqrt(x-1) dx. I have tried u substitution by using u=(x-1)^1/2 or using u=x-3. There has to be a trick to this problem that I'm not seeing and it's driving me nuts! This is a u substitution problem right? Any help is greatly appreciated!
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  2. #2
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    Re: U substitution indefinite integral

    Quote Originally Posted by Chewy View Post
    I have been trying to solve the indefinite integral (x-3)*sqrt(x-1) dx. I have tried u substitution by using u=(x-1)^1/2 or using u=x-3. There has to be a trick to this problem that I'm not seeing ...
    u = x-1

    du = dx

    x = u+1

    \int (x-3)\sqrt{x-1} \, dx

    substitute ...

    \int [(u+1)-3] \sqrt{u} \, du

    proceed ...
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  3. #3
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    Re: U substitution indefinite integral

    Hello, Chewy!

    \int (x-3)\sqrt{x-1}\,dx

    If the expression under the radical is linear,
    . . we can let u = the entire radical.

    Let u \:=\:\sqrt{x-1}\quad\Rightarrow\quad x-1 \:=\:u^2 \quad\Rightarrow\quad dx \:=\:2u\,du

    Substitute: . \int (u^2-2)\cdot u \cdot 2u\,du \;=\;2\int(u^4 - 2u^2)\,du

    . . . . . . . . . . =\;2\left(\tfrac{1}{5}u^5 - \tfrac{2}{3}u^3\right) + C \;=\; \tfrac{2}{15}u^3\left(3u^2-10)+C

    Back-substitute: . \tfrac{2}{15}\left(\sqrt{x-1}\right)^3\big[3(x-1) - 10\big] + C

    . . . . . . . . . . . =\;\tfrac{2}{15}(x-1)^{\frac{3}{2}}(3x-13) + C
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