We know, that given a function $f(t)$ where $t\geq 0$, its Laplace transform $F(s) = \mathcal{L}{\{f(t)\}}$ is defined as $F(s) = \mathcal{L}{\{f(t)\}} = \int_{0}^{\infty}{e^{-st}f(t)dt}$
I am given $f(t) = te^{-t}$ with $t\geq0$. So
$F(s) = \mathcal{L}{\{f(t)\}} = \lim_{t\to\infty}\int_{0}^{t}{e^{-st}te^{-t}dt}$
The expression within the integrand simplifies to
$\lim_{t\to\infty}\int_{0}^{t}{e^{-t-st}dt}$
Then I used integration by parts to yeild
$\lim_{t\to\infty}\frac{{e^{-s-1}t}}{(-s-1)(s+1)}$
Again, parts gives me another limit expressed as
$\lim_{t\to\infty}\frac{{te^{-s-1}t}}{(s+1)}$
I really do not think I am on the right track with this. Please, if someone would guide me in a more correct direction I can help other people too.
Please be careful using $t$ as a variable and as a limit. You want $$\lim_{T \to \infty} \int_0^T t\mathrm e^{-t}\cdot \mathrm e^{-st}~\mathrm dt$$
I get a different answer when I use integration by parts. I get $$\lim_{T \to \infty} \int_0^T t\mathrm e^{-t(s+1)}~\mathrm dt \ \ = \ \ \lim_{T\to \infty}\left[-\frac{st+t+1}{s^2+2s+1}\mathrm e^{-(s+1)t} \right]_0^T$$
I'll leave you to try to finish this. Remember to assume that $s>0$.