I was using Laplace to find solutions for $$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$$

with boundary conditions $$u(0,t)=1 \\ u(1,t)=1 \\ u(x,0)=1+ \sin \pi x$$

I used $U(x,s)= \mathcal L^{-1}{u(x,t)}$ and got $$U(x,s)=Ae^{x\sqrt s} + Be^{-x\sqrt s} + \frac{1}{s} + \frac{\sin \pi x}{s+\pi ^2}$$

$$u(x,t)= A\mathcal L ^{-1} \{e^{x\sqrt s} \} + B\mathcal L ^{-1} \{e^{-x\sqrt s} \} + 1 + e^{-\pi ^2t}\sin \pi x$$

How to solve for the first two inverse Laplace terms?