In a table and also on WolframAlpha, I stumbled upon this
http://www.wolframalpha.com/input/?i=laplace+transform+1%2F%281%2Bt%29
So the Laplace transform of $1/(1+t)$ is apparently $-e^s \text{Ei}(-s)$. There was no explanation of this $\text{Ei}(-s)$ function in neither the table nor WolframAlpha. So what is this weird function?
As you have already been informed by @MhenniBenghorbal, it is an exponential integral. $$ \mathcal{L}\Big\{\frac{1}{1+t}\Big\} = \int_0^{\infty}\frac{e^{-st}}{t + 1}dt $$ We can get the Laplace transform by letting $u = 1 + t$. Then $du = dt$ and we have \begin{align} \int_0^{\infty}\frac{e^{-st}}{t + 1}dt &= \int_1^{\infty}\frac{e^{-su + s}}{u}du\\ &= e^s\int_1^{\infty}\frac{e^{-su}}{u}du \end{align} where $\int_1^{\infty}\frac{e^{-su}}{u}du = -\text{Ei}(-s)$. Thus $$ \mathcal{L}\Big\{\frac{1}{1+t}\Big\} = -e^s\text{Ei}(-s). $$