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

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Answer 1 · 2016-01-20T20:35:38

#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.