Integrating $\frac{1}{u}$ Say I have $\int_{}{}\frac{\frac{1}{2}}{u + 1}du$
Why is $\frac{1}{2}\ln(u + 1) \neq \frac{1}{2}\ln(2u + 2)$?
Assuming I did this: 
$$\int_{}{}\frac{\frac{1}{2}}{u + 1}\ du = \int_{}{}\frac{1}{2u + 2}\ du$$

You are forgetting the constant for an indefinite integral.
Note that
$$\frac{1}{2} \int \frac{du}{u+1} = \frac{1}{2} \ln (u+1) + C,$$
and
$$\frac{1}{2} \ln(2u+2) = \frac{1}{2} \ln (u+1 ) + \frac{\ln 2}{2}.$$
