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    Integrate x(x-1)^1/2 dx, u=(x-1)^1/2

    so du/dx = 1/2(x-1)^-1/2
    dx/du=2(x-1)^1/2

    subbing u in: x*u*2(x-1)^1/2

    Rearranging u for x gives u^2 +1=x so subbing in x we get:

    (u^2+1)*(u)*2(x-1)^1/2
    so I've got (u^3+u)*2(x-1)^1/2... Im not sure how to get rid of that x now though... Am I going in the right direction? any help is greatly appreciated thanks.
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    Hi,

    I wouldn't use that substitution. Here's how I done it. Let u=x-1. Then du=dx. Now the Integral becomes integrate " (u+1)u^1/2 " du.
    where u=x-1. so root(x-1) is indeed u^1/2. Now expand the brackets and integrate as usual. You should arrive at 2/5(x-1)^5/2 + 2/3(x-1)3/2 + constant.

    Thanks Let me know if you need more help
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    (Original post by mathnerd2017)
    Hi,

    I wouldn't use that substitution. Here's how I done it. Let u=x-1. Then du=dx. Now the Integral becomes integrate " (u+1)u^1/2 " du.
    where u=x-1. so root(x-1) is indeed u^1/2. Now expand the brackets and integrate as usual. You should arrive at 2/5(x-1)^5/2 + 2/3(x-1)3/2 + constant.

    Thanks Let me know if you need more help
    I know that substitution isnt a good choice but its what the question told me to use

    And the answer they said is 2/5(3x+2)(x-1)^3/2

    They literally have a whole bunch of questions using the squareroot substitution.
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    [QUOTE=MrToodles4;74418448]I know that substitution isnt a good choice but its what the question told me to use

    And the answer they said is 2/5(3x+2)(x-1)^3/2

    They literally have a whole bunch of questions using the squareroot substitution.[/QUOTE

    Yes, I have solved it using your substitution. The method is as follows. u=(x-1)^1/2.
    After a little work, I arrive at 2(x-1)^1/2du=dx

    Sub that in. We arrive at integrate "(1+u^2)2u^2" where u^2=x-1.

    Then expand out the integral and perform usual integration.

    Hope that helps!!
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    (Original post by MrToodles4)
    I know that substitution isnt a good choice but its what the question told me to use

    And the answer they said is 2/5(3x+2)(x-1)^3/2

    They literally have a whole bunch of questions using the squareroot substitution.
    Also, what I wrote is correct, it just requires a little algebra. Manipulate it a bit, get it in a single fraction and you arrive at the required answer.
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    [QUOTE=mathnerd2017;74418572]
    (Original post by MrToodles4)
    I know that substitution isnt a good choice but its what the question told me to use

    And the answer they said is 2/5(3x+2)(x-1)^3/2

    They literally have a whole bunch of questions using the squareroot substitution.[/QUOTE

    Yes, I have solved it using your substitution. The method is as follows. u=(x-1)^1/2.
    After a little work, I arrive at 2(x-1)^1/2du=dx

    Sub that in. We arrive at integrate "(1+u^2)2u^2" where u^2=x-1.

    Then expand out the integral and perform usual integration.

    Hope that helps!!
    Many thanks.
 
 
 
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