As I already told you, graph it. Did you do that?

Note what happens near x = 0 & 1. And 1/[log(x-x^(2))] is bounded:

0≤ 1/[log(x-x^(2))] < 55, say.

Since its bounded (below & above) * continuous everywhere in [0,1] except at a few points (namely, 0 & 1), the integral thus gives a value, hence your conclusion.

lim x-->0 1/ln(x-x^2) = 0

lim x-->1 1/ln(x-x^2) = 0

f(x) = 1/ln(x-x^2)

f(0) = 0

f(1) = 0

http://www.wolframalpha.com/input/?i=1%2Fln(x-x%5E...

http://www.wolframalpha.com/input/?i=1%2Fln(x-x%5E...

Thus f(x) is continuous on [0,1]