We can consider two integrals:
∫01dt/(t(1-ln(t))2+∫1∞dt/(t(1+ln(t))2=∫dt/(t(1+|ln(t)|)2, and let J be the first integral and the second K.
because between t=0 and t=1, ln(t) is negative so |ln(t)|=-ln(t)=ln(1/t); and in the interval [1,∞) |ln(t)|=ln(t).
Let x=1-ln(t), then -dx=dt/t; when t=1, x=1 and when t=0, x→∞, so the limits for J are x∈(-∞,1].
J=-∫∞1dx/x2=[1/x]∞1=1. Or we could revert to t: J=[1/(1-ln(t))]01=1.
Let x=1+ln(t), then dx=dt/t; when t=1, x=1 and when t→∞, x→∞, so the limits for K are the same as for t.
K=∫1∞dx/x2=[-1/x]1∞. Reverting to t: K=[-1/(1+ln(t))]1∞=1.
So J+K=2. It would appear that the integral has become a definite integral, even though no limits were specified for the original integral.
However, see the graph below. The integral is the total area beneath the curve, which approaches asymptotically the y-axis on the left and the x-axis on the right, enclosing what seems to be a finite area.