For the integral:

$\displaystyle

\int_{0}^{\frac{1}{2}}{\frac{1}{x\ln^2{x}}}dx

$

Using the substitution:

$\displaystyle

u=\ln{x}

$

$\displaystyle

du=\frac{1}{x}

$

I am getting the following for my new terminals:

$\displaystyle

-\ln{2}

$

$\displaystyle -\infty$

To get my new terminals I have simply plugged the values of my old terminals into my u substitution (u=lnx).

When I evaluate the integral I wind up with a limit that needs to be evaluated, as follows:

$\displaystyle \lim_{t\rightarrow-\infty}\frac{-1}{ln(-ln(2)}+\frac{1}{ln(t)}$

When I evaluate that I come up with:

$\displaystyle \frac{1}{ln(-ln(2))}$

Have I done something wrong here?