hey uhh.. my teacher never taught me how to differentiate log/ln functions or integrate them.. so i am asking, what is the integral of $\displaystyle ln(t^2 - 3t + 3)$
I'd suggest to start by using integration by parts:
$\displaystyle \int u \, dv = uv - \int v \, du$.
Let $\displaystyle u = \ln (t^2 - 3t + 3) \Rightarrow du = \frac{2t - 3}{t^2 - 3t + 3} \, dt$ and $\displaystyle dv = dt \Rightarrow v = t$.
Now you have some spade work to do.