
Improper Integral
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?

Your notation is a little confused for the limit. However, 1/ln(2) is the correct answer. Certainly, your setup looks fine.

I tried to correct that last line, but somehow wound up making creating a double thread. I am not coming up with 1/ln(2). I am coming up with 1/ln(ln(2)). I know that isn't right.
Can you tell me where I am going wrong?
Oh... and I did mean to say "add up the fractions" not "add up the constants".

Check your antiderivative.

I had the following:
$\displaystyle \int{\frac{du}{u^2}}$
Which gave me:
$\displaystyle \frac{1}{u}$
I try to find my antiderivatives and verify them with my TI89. In this case, the 89 said I was correct.

The derivative of $\displaystyle \ln x$ is $\displaystyle \frac{1}{x}$ so that is...
$\displaystyle \displaystyle \int \frac{dx}{x\ \ln^{2} x} = \frac{1}{\ln x} + c$ (1)
... and ...
$\displaystyle \displaystyle \int_{0}^{\frac{1}{2}} \frac{dx}{x\ \ln^{2} x} = \frac{1}{\ln 2}$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$

Under your substitution your integral becomes
$\displaystyle \int \limits_{\infty}^{\ln 2} \dfrac{du}{u^2}$
and so
$\displaystyle \lim \limits_{a \to  \infty} \int \limits_{a}^{\ln 2} \dfrac{du}{u^2} = \lim \limits_{a \to  \infty} \dfrac{1}{u}\left. \right \limits_{a}^{\ln 2} = \lim \limits_{a \to  \infty} \dfrac{1}{\ln 2}+ \dfrac{1}{a}$