Finding Laplace Transform of $te^{-t}$ I started with this integral: 
$$ \int_{0}^{\infty} e^{-st}\cdot te^{-t}dt$$
= $$\int_{0}^{\infty} te^{-(s+1)t}dt$$
let $dv=e^{-(s+1)t}dt, u=t$ and thus $v=-\frac{1}{s+1}e^{-(s+1)t}dt, du=dt$
$\rightarrow$
$-\frac{t}{s+1}e^{-(s+1)t}|_0^\infty + \frac{1}{s+1}\int_{0}^{\infty}e^{-(s+1)t}dt$
= $-\frac{1}{(s+1)^2}e^{-(s+1)t}|_0^\infty$ 
What mistakes have I made or do I just use L'Hopital's rule to finish?

As an exercise I would have my DE students see how many of the elementary Laplace transform rules they could derive using only linearity plus $L\{y^\prime\}=sL\{y\}-y(0)$. 
It is simple then to find $L\{e^{at}\}=\frac{1}{s-a}$. Having that one can find $L\{te^{at}\}$ as follows:
Let $y=te^{at}$. Then $y^\prime=e^{at}+ate^{at}$.
\begin{eqnarray}
L\{y^\prime\}&=sL\{y\}-y(0)\\
L\{e^{at}+ate^{at}\}&=sL\{te^{at}\}\\
L\{e^{at}\}+aL\{te^{at}\}&=sL\{te^{at}\}\\
\frac{1}{s-a}&=(s-a)L\{te^{at}\}\\
L\{te^{at}\}&=\frac{1}{(s-a)^2}
\end{eqnarray}
