How do you integrate dx/x(lnx)^3?

Question

How do you integrate dx/x(lnx)^3?

Answer 1

INTEGRAL[(lnx)^3 * dx/x]

let u = ln x, du = dx/x

INTEGRAL[(u)^3 * du] = (u^4)/4 + C =  ( (lnx)^4)/4 + C

Answer 2

∫ dx / [x (lnx)^3] =

rearrange the integrand as:

∫ (lnx)^(-3)  (1/x) dx =

note that the integrand includes both the function lnx (even if powered)

and its derivative;

thus let lnx = u

differentiate both sides:

d(lnx) = du →

(1/x) dx = du

thus, substituting, you get:

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

[u^(-3+1)] /(-3+1) + C =  [u^(-2)] /(-2) + C =

[-1/(2 u^2)] + C

thus, substituting back u = lnx, you get:

∫ dx / [x (lnx)^3] = {-1/ [2 (lnx)^2]} + C

I hope it helps..

Bye!