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Math Help - need help with intergration

  1. #1
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    need help with intergration

    I'm stuck on intergrating a function. the function is (x-2)/(x^2-4x+5)^.5. The problem states that we are to use (x-2)^2 as u for subsition. So i took the derivative of (x-2)^2 and plugged du into the intergal. Then I'm stuck. I have .5*the intergal of du/(u+1)^.5. I don't know what to do after that. Any help?
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by flash9286 View Post
    I'm stuck on intergrating a function. the function is (x-2)/(x^2-4x+5)^.5. The problem states that we are to use (x-2)^2 as u for subsition. So i took the derivative of (x-2)^2 and plugged du into the intergal. Then I'm stuck. I have .5*the intergal of du/(u+1)^.5. I don't know what to do after that. Any help?
    You took care of hardest part!

    .5*INT 1/sqrt(u + 1) du

    let u + 1 = v --> dv = du

    .5*INT 1/sqrt(v) dv
    .5*INT (v)^(-1/2)
    .5*2*(v)^(1/2) + C
    = sqrt(v) + C
    = sqrt(u + 1) + C
    = sqrt[(x - 2)^2 + 1] + C
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  3. #3
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    Thanks for your help.
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  4. #4
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    Hello, flash!

    They surely suggested that substitution to give you practice,
    . . but there is a simpler way . . .

    . . . . . . . . . . .(x - 2) dx
    We have: . ∫ --------------- . = .
    (x - 4x + 5)^{-} (x - 2) dx
    . . . . . . . . . √x - 4x + 5


    Let: .u .= .x - 4x + 5 . . . . du .= .(2x - 4)dx . . . . dx .= .du/2(x-2)


    Substitute and we have: .
    u^{-} du

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