Find the causal inverse transform, for t ≥ 0. By linearity, invert each partial fraction

The inverse Laplace formula for $\frac{1}{{(s-a)}^n}$ is $t^{n-1}·\frac{\exp \left(at\right)}{(n-1)!}$; it gives for $\frac{1}{{(s+1)}^3}$: $\frac{t^2\exp (-1·t)}{2!}$

⇥ⓘ Multiplying by -1 gives the opposite: $-1·t$ = $-t$■The factorial of a number is the product of all the integers from 1 to that number: $2!$ = $2$

⇥ $\frac{t^2\exp (-t)}{2}$

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base: $\frac{t^2\exp (-t)}{2}$ = $\frac{t^2}{2\exp \left(t\right)}$

▶Inverse Laplace transform:   $\frac{t^2}{2\exp \left(t\right)}$