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    All the other questions seemed so easy

    int. between limits 5 - 1 of x/(2x-1)^(3/2) dx

    It doesnt tell you what to substitute, so I opted for u^(2/3)=2x-1...

    So to make the dU/dX i had to integrate implicitly... erm yeah... then it gets really weird numbers and I get the wrong answer; any help would be appreciated
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    can you post the answer please? I just don't want to post an incorrect solution and frighten you
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    4/3
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    (Original post by Talwin)
    All the other questions seemed so easy

    int. between limits 5 - 1 of x/(2x-1)^(3/2) dx

    It doesnt tell you what to substitute, so I opted for u^(2/3)=2x-1...

    So to make the dU/dX i had to integrate implicitly... erm yeah... then it gets really weird numbers and I get the wrong answer; any help would be appreciated

    Let u=2x-1

    So you are left with 1/4int.sqrtu+1/4int.u^-3/2 (which requires only P1 knowledge to integrate).
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    Thankyou, but how did you get from int. (1 + u)/u^(3/2) du to 1/4int.sqrtu+1/4int.u^-3/2 ? ... have I gone wrong somewhere?
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    ∫x/(2x-1)^(3/2)dx
    let u = 2x-1
    so x = (u +1)/2

    du/dx = 2 so dx = du/2

    (1/2)∫(u+1)/u^(3/2) du/2

    take out the half:

    (1/4)∫(u^(-1/2) + u^(-3/2) du

    giving:

    (1/4)[2.u^(1/2) - 2.u^(-1/2)]

    =(1/2)[(2x-1)^(1/2) - (2x-1)^(-1/2)] and this is between 1 and 5

    so it equals:

    (1/2)[(3 - 1/3) - (1 - 1)]

    (1/2)[8/3]

    = 4/3

    Hope this helps.
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    thanks makes sensee...no...

    OMG!! MY BROTHER JUST STROLLED INTO MY ROOM IN A DRESS AND MAKE UP!!! :eek:
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    yeah... it happened, but i will avoid going off topic
 
 
 

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