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    Sorry! I'm just struggling with a few questions, I've tried doing them like 10 times and I seem to be getting a close, but not the same answer. So if anyone could do even just one, and do each step for me I'd totally love you forever =D xxx

    ∫ 5x / 5x-2 dx u = 5x-2

    ∫ x / (2x-1)3dx u = 2x-1

    ∫ x / √(x2-1)
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    (Original post by -x-buttercup-x-)
    Sorry! I'm just struggling with a few questions, I've tried doing them like 10 times and I seem to be getting a close, but not the same answer. So if anyone could do even just one, and do each step for me I'd totally love you forever =D xxx

    ∫ 5x / 5x-2 dx u = 5x-2

    ∫ x / (2x-1)3dx u = 2x-1

    ∫ x / √(x2-1)

    first one:

    u = 5x-2
    du/dx = 5
    dx/du = 1/5
    dx = du/5

    ∫ 5x / 5x-2 dx
    = (1/5)∫ ((u+2)/u) du
    = (1/5)∫(1+2/u) du
    = (1/5)(u+2lnu)+c
    = (5x-2)/5 + (2/5)ln(5x-2) + c
    = x + (2/5)ln(5x-2) + k (k = c-2/5)

    Im not really sure with the second one, the 3 is confusing me is it ∫ (x / 3(2x-1))dx ?
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    ∫ x / √(x2-1)
    ∫ x / √(x²-1) dx if thats what you ment then let
    u=x²-1 => x²=u+1 => x=(u+1)1/2
    du/dx=2x = 2(u+1)1/2
    dx=du/2(u+1)1/2

    ∫ (u+1)1/2 / (u1/2)2(u+1)1/2 du
    =∫ 1/(2u1/2) du
    =∫(1/2)u-1/2 du
    =u1/2+c
    =(x²-1)1/2 +c
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    Thankyou both of you =D You'll be getting pos rep shortly =D xxx
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    oh and yep I did mean x squared, it did come up right in word, but most of gone funny when I pasted it into here sorry x
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    (Original post by -x-buttercup-x-)
    ∫ x / (2x-1)3dx u = 2x-1
    Assuming you mean ^3
    x =(u+1)/2
    dx/du = u/2

    1/4 ∫ (u+1)u/u3 du
    1/4 ∫(u+1)/u2 du
    1/4 ∫1/u + 1/u2 du
    1/4 [ lnu - 1/u ] + C
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    (Original post by YYYY)
    Assuming you mean ^3
    x =(u+1)/2
    dx/du = u/2

    1/4 ∫ (u+1)u/u3 du
    1/4 ∫(u+1)/u2 du
    1/4 ∫1/u + 1/u2 du
    1/4 [ lnu - 1/u ] + C
    dx/du = 1/2 surely?

    1/4 ∫ (u+1)/u3 du
    1/4 ∫1/u2 + 1/u3 du
    -1/4[1/3u-3 + 1/4u-4 + c]

    -----------------------------------------------------

    EDIT: MY METHOD WAS WRONG FOR THIS ONE. you can use x = sec u, but i can't be bothered to go through it. if you have a formula book, it's nice and simple, otherwise i'd use the substitution shown earlier.
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    (Original post by sea tea)
    dx/du = 1/2 surely?

    1/4 ∫ (u+1)/u3 du
    1/4 ∫1/u2 + 1/u3 du
    -1/4[1/3u-3 + 1/4u-4 + c]
    ****!
    My mistake
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    Awww thanks again, I actually understand it now.I will rep you all as soon as I can
 
 
 
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