Question

What is the antiderivative of `1/[x(ln(x^3))]`?

Answer 1

`int 1/(xln(x^3)) dx`

Note that `d/dx(lnx) = 1/x` and,

more importantly,

`d/dx(ln(x^3)) = 3/x`.
(You can get this by the chain rule, or more simply, by noting that `ln(x^3) = 3lnx`.)

This integral can be evaluated by substitution:

Let `u = ln(x^3)`, then `du = 3/x dx`, so `1/x dx = 1/3 du`

Upon substitution, the integral becomes:

`int 1/3 u^-1 du = 1/3 ln abs u+C`.

Therefore:

`int 1/(xln(x^3)) dx = 1/3 ln abs ln(x^3)+C`

Another way to proceed:

`int 1/(xln(x^3)) dx = int 1/(3xln(x)) dx`

`= 1/3 int 1/(xln(x)) dx`

Let `u = lnx` and proceed to get:

`int 1/(xln(x^3)) dx = 1/3 lnabs(lnx) +C`

It looks different, but what is the difference?

`1/3 ln abs ln(x^3) =1/3 ln abs(3lnx)`

`= 1/3(ln3 + ln abs lnx)+C`

`= 1/3ln3 + 1/3 ln abs lnx+C`

So the difference between the expressions is a constant.
The two general answers simply have different `C`'s for particular answers.