Finding the inverse Laplace transform of
$L^{-1}\left( \dfrac {1}{\left( x+1\right) ^{2}}\right)$
I understnd that the inverse laplace of
$L^{-1}\left(\dfrac {1}{\left( x+1\right) }\right)$
Is equal to the negative exponent of x
$e^{-x}$
But cant seem to see the intuition into finding the inverse laplace function of the function I have given
Thanks
\begin{align} \mathcal L\left\{t^{n} e^{-\alpha t} \cdot u(t)\right\}&=\int_0^\infty e^{-st}t^{n} e^{-\alpha t}\mathrm d t=\int_0^\infty e^{-(s+\alpha)t}t^{n} \mathrm d t\\ &=\frac{1}{(s+\alpha)^{n+1}}\int_0^\infty e^{-u}u^{n} \mathrm d u=\frac{1}{(s+\alpha)^{n+1}}\Gamma(n+1)\\ &=\frac{n!}{(s+\alpha)^{n+1}} \end{align} So you have that $$ \mathcal L^{-1}\left\{\frac{1}{(x+1)^{2}}\right\}=t e^{-t} \cdot u(t) $$
Another way is $$ \mathcal L\left\{e^{-\alpha t} \cdot u(t)\right\}=\frac{1}{s+\alpha} $$ and using the property $ (-1)^{n} F^{(n)}(s) =\mathcal L\left\{t^n f(t)\right\}$ we have $$\mathcal L\left\{t e^{-\alpha t} \cdot u(t)\right\}=(-1)\left(\frac{1}{s+\alpha}\right)'=\frac{1}{(s+\alpha)^2}$$