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?

$$\mathcal{L\{te^{-t}\}}=\int^{\infty}_{0} e^{-st}te^{−t}dt $$
$$=\int^{\infty}_{0} e^{-t(s+1)}tdt $$
substitute 
$$st=x;dt=\frac{dx}{s}$$
$$=\frac{1}{s^2}\int^{\infty}_{0} xe^{-\frac{x}{s}(s+1)}dx $$
$\frac{x(s+1)}{s}=m;dx\frac{(s+1)}{s}=dm;
$
$$=\frac{1}{(s+1)^2}\int^{\infty}_{0}m \ e^{-m}dm $$
$$=\frac{\Gamma (2)}{(s+1)^2}$$
$$=\frac{1}{(s+1)^2}$$
