Inverse Laplace Transform of s/(s+1)

HINT: $$\frac s{s+1}=1-\frac1{s+1}$$

We know, $\mathcal{L}(e^{at})=\frac1{s-a}$

and we can prove $\mathcal{L}(\delta(t-b))=e^{-bs}$, put $b=0$

Simple way to solve this problem sorry about the formatting

\begin{align} &=\frac s{s+a}\\ &=\frac{(s+a)-a}{s+a}\\ &=\frac{s+a}{s+a}-\frac a{s+a}\\ &=1-\frac a{s+a}\\ &=δ(t)-\mathcal(ae^{-at}) \end{align}