Help me complete this Differential Geometry equation for a geodesic.

Question

I am wondering what happens to the geodesic equation if there is a constant acceleration,$A_3$: $$\frac{d^2x^\nu}{ds^2}=A_3$$ $$\frac{dx^\nu}{ds}=\frac{\partial x^\nu}{\partial x^\beta}\frac{d x^\beta}{ds}$$ $$\frac{d}{ds}\left[\frac{dx^\nu}{ds}\right]=\frac{d}{ds}\left[\frac{\partial x^\nu}{\partial x^\beta}\frac{d x^\beta}{ds}\right]=A_3$$ $$\frac{d}{ds}\left(\frac{dx^\nu}{d x^\beta}\right)=\frac{\partial}{\partial x^\beta}\left(\frac{dx^\nu}{ds}\right)=\frac{\partial^2 x^\nu}{\partial x^\beta \partial x^\alpha}\frac{dx^\alpha}{ds}$$ Using the product rule, we get: $$\frac{d}{ds}\left[\frac{\partial x^\nu}{\partial x^\beta}\frac{d x^\beta}{ds}\right]=\frac{\partial x^\nu}{\partial x^\beta}\frac{d^2 x^\beta}{ds^2}+\frac{\partial^2 x^\nu}{\partial x^\beta \partial x^\alpha}\frac{dx^\alpha}{ds}\frac{d x^\beta}{ds}=A_3$$ This is the part where I need help. I want to isolate the acceleration term $\frac{d^2 x^\beta}{ds^2}$ and construct a Christoff symbol, so I multiply each side by the inverse of the transform matrix: $\frac{\partial x^\nu}{\partial x^\beta}^{-1}$ $$\frac{d^2 x^\mu}{ds^2}+\left[\left(\frac{\partial x^\nu}{\partial x^\mu}\right)^{-1}\frac{\partial^2 x^\nu}{\partial x^\beta \partial x^\alpha}\right]\frac{dx^\alpha}{ds}\frac{d x^\beta}{ds}=\left(\frac{\partial x^\nu}{\partial x^\mu}\right)^{-1}A_3$$ $$\frac{d^2 x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{d x^\beta}{ds}=\left(\frac{\partial x^\nu}{\partial x^\mu}\right)^{-1}A_3$$ But I'm pretty sure I can't use the same indices when I perform this last operation and the inverse matrix notation just doesn't look right. Please help me complete this formula with the proper notation (or confirm that the notation is correct).