Calculus

help!

∫((ln x^2)/x)) dx

u= ln x^2
du= 2 ln x dx (?)

Not sure how to set this one up?

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asked by Steff
  1. let u = ln(x^2) or 2lnx
    then du/dx = 2/x
    du = (2/x)dx

    let dv = (1/x) dx
    dv/dx = 1/x
    v = lnx

    then
    ∫ u dv = uv - ∫v du
    ∫2lnx(1/x) dx = 2lnx(lnx) - ∫lnx(2/x)dx
    2 ∫2lnx / x dx = 2 (lnx)^2
    ∫(ln x^2)/x dx = (lnx)^2


    proof:
    d (lnx)^2) /dx = 2(lnx)(1/x) = ln (x^2) / x

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    posted by Reiny
  2. Thank you!

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    posted by Steff

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