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

Question

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

Answer 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

Answer 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!

Answer 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