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Quick P3 Integration using Substitution qu. watch

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    Question.. "using u^2 = x - 1, find ∫(1 / (1 + √(x-1))) .dx"

    so 2u . du = 1
    du = 1/(2u)

    therefore ∫(1 / (1 + √(x-1))).dx = ∫(1 / (1 + u)) .2u .du

    ∫(2u / (1 + u)) .du = ??! looks so easy i seem to get lost!!


    Cheers, SAHIR
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    (Original post by Sahir)
    Question.. "using u^2 = x - 1, find ∫(1 / (1 + √(x-1))) .dx"

    so 2u . du = 1
    du = 1/(2u)

    therefore ∫(1 / (1 + √(x-1))).dx = ∫(1 / (1 + u)) .2u .du

    ∫(2u / (1 + u)) .du = ??! looks so easy i seem to get lost!!


    Cheers, SAHIR
    ∫(u / (1 + u)) .du = ∫ ((u+1) - 1) /(1+u) = ∫ 1 - (1/(1+u)) du = u - ln|u+1|.
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    2u.du = 1.dx, so you replace the dx with 2u.du, not 1/2u .du
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    duuuuh.. i feel so stooopid!

    Heh, cheers theone... rapid response maths at its best.
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    (Original post by elpaw)
    2u.du = 1.dx, so you replace the dx with 2u.du, not 1/2u .du
    Yeah that was just a lapse in concentration when i typed it.. wrote down it correctly and its correct in the next line.. cheers though elpaw.
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    (Original post by theone)
    ∫(u / (1 + u)) .du = ∫ ((u+1) - 1) /(1+u) = ∫ 1 - (1/(1+u)) du = u - ln|u+1|.
    + C!

    and then you have to substitute x back in.
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    (Original post by elpaw)
    + C!

    and then you have to substitute x back in.
    Always there to pick me up when I fall
 
 
 
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