Transformada inversa de Laplace 1/s+5?

=h(t)+5d(t)

Pela tabela de Transformadas de Laplace

$1/(s+a) = e^(-at)$1/(s+a)=e(−at) , nesse caso a=5

$1/(s+5) = e^(-5t)$1/(s+5)=e(−5t)