Question:

How do I integrate 1/(x*(ln(x^3)))?

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How do I integrate 1/(x*(ln(x^3)))?

I'm a little confused because I think I'm supposed to use substitution, but I can't get the resulting x^2 to make sense

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  1. your suppose to use the product rule and the chain rule in this

    first seperate it by

    1/x(1/ln(x^3))

    so by product rule it's

    -1/x^2(ln(x^3)) + 1/x(-1/[ln(x^3)^2)])(1/x^3)(3x^2)

    = -1/x^2(ln(x^3)) - 3x^2 / [x^4(ln[x^3]^2)]

    ok so tht was my answer but this is what i got from online integrator

    Ln[Ln[x^3]]/3

    integrals.com

    so mine might be wrong


  2. ∫ {1 / [x ln(x³)]} dx =

    first, you have to recall the property of logarithms log a^(n) = n log a,

    therefore:

    ∫ {1 / [x ln(x³)]} dx = ∫ {1 / [x (3 lnx)]} dx =

    taking the constant out:

    (1/3) ∫ (1 /lnx) (1/x) dx =

    note that your integrand includes both the function lnx and its derivative (1/x),

    thus

    let lnx = u

    differentiate both sides as:

    d(lnx) = du →

    (1/x) dx = du

    thus, substitute, yielding:

    (1/3) ∫ (1 /lnx) (1/x) dx = (1/3) ∫ (1 /u) du =

    (1/3) ln | u | + C

    then, substituting back u = lnx, you get:

    ∫ {1 / [x ln(x³)]} dx = (1/3) ln | lnx | + C

    I hope it helps..

    Bye!

  3. note: ln(x^3) = 3 ln x

    thus .. . .

    we have

    1/3 ∫ 1/[x ln x] dx

    let u = ln x

    then du = 1/x dx

    we now have

    1/3 ∫ 1/u du

    = 1/3 ln|u|

    = 1/3 ln|ln x| + C

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