How do you evaluate the integral #int 1/(x(lnx)^2) dx# from 0 to 1 if it converges?

1 Answer
Nov 29, 2016

The definite integral is not convergent.

Explanation:

First solve the indefinite integral using the substitution:

#x=e^t#
#dx=e^tdt#

#int dx/(x(lnx)^2) = int (e^tdt)/(e^t t^2) = int(dt)/t^2 = -1/(3t^3)=-1/(3(ln x)^3)#

The primitive is defined for #x in (0,1) # but:

#lim_(x->0^+) -1/(3(ln x)^3) = 0#

#lim_(x->1^-) -1/(3(ln x)^3) = +oo#

So the definite integral does not converge.