Solve using Laplace transforms

I want to solve the PDE
$ $ x\frac{\partial u}{\partial t} +\frac{\partial u}{\partial x}=x$ $ $ $ u(x,0)=0, x>0$ $ $ $ u(0,t)=0, t>0$ $ for $ x>0,t>0$ using a Laplace transform in t.
I Laplace transform the equation and get $ $ xs\hat{u}(x,s)+\frac{\partial}{\partial x}\hat{u}(x,s)=\frac{x}{s}$ $ where $ \hat{u}(x,s)=L(u(x,t))=\int_{t=0}^{\infty} e^{-st}u(x,t)dt$ .
Solving this ODE, I obtain $ $ \hat{u}(x,s)=\frac{1}{s^2} + A(s)e^{\frac{-sx^2}{2}}.$ $ I then apply the second boundary condition to get $ $ \hat{u}(x,s)=\frac{1}{s^2}-\frac{1}{s^2}e^{\frac{-sx^2}{2}}.$ $ Now I need to inverse transform to get $ u(x,t)$ . I know that $ $ L^{-1}(\frac{1}{s^2})=t$ $ but I don’t know how to find $ $ L^{-1}(\frac{1}{s^2}e^{\frac{-sx^2}{2}}).$ $