Integration by Parts using Ln(x)

[FallingMan]

Homework Statement

\int_4^5 \frac{dx}{(3x)(ln(x))(ln^3(ln(x)))} \,

Homework Equations

I'll probably need to use u-substitution as well as integration by parts for this problem.

The Attempt at a Solution

\int_4^5 \frac{dx}{(3x)(ln(x))(ln^3(ln(x)))} \,

1. Factor out 1/3 for convenience.

\frac{1}{3}\, \int_4^5 \frac{dx}{(x)(ln(x))(ln^3(ln(x)))} \,


2. Let u = ln(x), du = 1/x * dx, thus dx = du * x


\frac{1}{3}\, \int_4^5 \frac{(x)(du)}{(x)(u)(ln^3(u))} \, = \frac{1}{3}\, \int_4^5 \frac{du}{(u)(ln^3(u))} \,

From integration by parts we know that ∫g*dv = g*v - ∫dg*v

I'm stuck here with regards to what to assign as g and what to assign as dv.

I could make:
g = 1/u
dg = -1/u^2

dv = 1/ln^3(u) * du
v = ?

I'm not sure how to integrate dv to get to v to make the integration by parts work.

Any help would be appreciated.

 

[HomogenousCow]

Make a second substitution

 

[lurflurf]

Integration by parts is not helpful here

You wither want two use the substitution

u=(log log x)^3

or equivalently another substitution after yours

u=log x
w=(log u)^3

 

[HomogenousCow]

That second substitution complicates the problem, a simple w=lnu is enough

 

[FallingMan]

Thanks for your help, guys. I think I understand what you're saying.

Let g = ln(u), dg = 1/u * du, thus du = dg* u

Substituting back in I get:\frac{1}{3}\, \int_4^5 \frac{dg*u}{(u)(g^3)} \, = \frac{1}{3}\, \int_4^5 \frac{dg}{g^3} \, = -\frac{1}{2g^2}\,\frac{1}{3}\,

and then I would substitute back to reobtain x and apply the limit from 4 to 5.

Thank you, guys. Much appreciated.