Inverse Laplace transform of $\frac{1}{e^{-s}+1}$

Question

I would like to know how to perform the following inverse Laplace transform: \begin{equation} \mathcal{L}^{-1}\left\{\frac{1}{ e^{-s} + 1}\right\}\,. \end{equation} This expression somehow relates the dc armature current with motor speed in rad/sec.

Answer 1

$$ \frac{1}{-s+\left(e+1\right)}=-\frac{1}{s-\left(e+1\right)}\Rightarrow\mathcal{L}^{-1}\left(-\frac{1}{s-\left(e+1\right)}\right)=-e^{\left(e+1\right)t}u\left(t\right) $$

Answer 2

It is well known that $F(s-a)=L(e^{at}f(t))$ and hence $ L^{-1}(F(s-a))=e^{at}f(t)$, Setting $a=1, F(s)=e^{s}$ we obtain $L^{-1}(e^{s-1})=e^{t}L^{-1}(e^{s})$.But $L(\delta (t-a))=\int_{0}^{\infty }e^{-st}\delta(t-a)dt$ and since $\delta(t-a)$$ =0 $ except at $a$ we assume that the integral is $e^{-sa}\int_{0}^{\infty }\delta (t-a)dt.$ And by definition of the delta function $\int_{0}^{\infty }\delta (t-a)dt$=$1$ therefore $L(\delta (t-a))=e^{-sa}$ and $L(\delta (t+1))=e^{-(-1)s}=e^{s}$. Thus $L^{-1}(e^{s})=\delta (t+1)$ and the answer to our problem is $e^{t}\delta (t+1)$.! I must admit delta function is tricky!