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?

If you see an integral of the form $$\int t f(t) \,dt$$ then try partial integration! 
Assuming $s\neq 1$
\begin{align}\int_0^{\infty} t f(t) \,dt &= \left[t\frac{-1}{s+1}e^{-(s+1)t} \right]_{t=0}^\infty-\int_0^{\infty} \frac{-1}{s+1}e^{-(s+1)t} \,dt \\&= 0 - 0 + \left[ \frac{-1}{(s+1)^2}e^{-(s+1)t} \right]_{t=0}^\infty \\ &= \frac{1}{(s+1)^2}\end{align}
