Integrate 1/(xlnxln(lnx))?

Question

Thankyou

Answer 1

∫ dx / {x lnx [ln(lnx)} =

∫ {1 / {lnx [ln(lnx)} } (1/x) dx =

note that (fortunately) the integrand includes both the function lnx and its

derivative (1/x);

thus, let  lnx = u

differentiate both sides into:

d(lnx) = du →

(1/x) dx = du

thus, substituting, you get:

∫ {1 / {lnx [ln(lnx)} } (1/x) dx = ∫ {1 / [u (ln u)] } du =

∫ [1 /(ln u)] (1/u) du =

similarly, you have both the function ln u and its derivative (1/u),

thus let ln u = t

differentiate both sides:

d(ln u) = dt →

(1/u) du = dt

then, substituting again, you get:

∫ [1 /(ln u)] (1/u) du = ∫ (1/ t) dt = ln | t | + C

then being t = ln u and u = lnx, substituting back you get:

∫ dx / {x lnx [ln(lnx)} = ln | ln(lnx) | + C

I hope it helps...

Bye!

Answer 2

∫dx/[xlnx*ln(lnx)]=

dx/x=d(lnx)

=∫d(lnx)/[lnx*ln(lnx)=

d(lnx)/lnx=d[ln(lnx)]

=d[ln(lnx)]/ln(lnx)=

=d{ln[ln(lnx)]}+c