simple laplace transformation Why do:


*

*$L(5t)=\frac{5}{s^2}$ and not $\frac{5}{s}$?

*$L(te^{-2t})=?$
I can work with laplace transformation table 
 A: $$\mathcal{L}(5t)(s) = \int_0^{+\infty} 5t\ e^{-st}\ \text{d}t = \frac{5}{s^2}$$
$$\mathcal{L}(te^{-2t})(s) = \int_0^{+\infty} t\ e^{-2t}\ e^{-st}\ \text{d}t = \frac{1}{(s+2)^2}$$
Integrals are trivial. If you reached this level of theory, you can compute them without problems.
Important LT properties here
http://lpsa.swarthmore.edu/LaplaceZTable/LaplacePropTable.html
A: Since you say you can work with transform tables, rather than deriving from the definition I will use two common results from the tables.
Using $\mathcal{L}\{y^\prime\}=s\mathcal{L}\{y\}-y(0)$ let


*

*$y_1=5$

*$y_2=5t$


$\mathcal{L}\{y_1\prime\}=s\mathcal{L}\{y_1\}-y_1(0)$ therefore
$\mathcal{L}\{0\}=s\mathcal{L}\{5\}-5$ so $\mathcal{L}\{5\}=\dfrac{5}{s}$
$\mathcal{L}\{y_2^\prime\}=s\mathcal{L}\{y_2\}-y_2(0)$ therefore
$\mathcal{L}\{5\}=s\mathcal{L}\{5t\}-5(0)$. Thus
$\dfrac{5}{s}=s\mathcal{L}\{5t\}$ so
$\mathcal{L}\{5t\}=\dfrac{5}{s^2}$.
Using $\mathcal{L}\{ty\}=-\dfrac{d}{dy}\mathcal{L}\{y\}$
$\mathcal{L}\{te^{-2t}\}=-\dfrac{d}{dy}\left(\dfrac{1}{s+2}\right)=\dfrac{1}{(s+2)^2}$
