Your mistake is in your conservation energy equation. The way you wrote it is valid only when falling from infinity, from rest. The correct is:
$$dE=dK+dU=0,$$
that is
$$mvdv=-\frac{K}{x^2}dx,$$
where $K\equiv Gm_1m_2$. Integrating from $(x_i,v_i)$ to $(x,v)$ we get
$$\frac 12m(v^2-v_i^2)=K\left(\frac{1}{x}-\frac{1}{x_i} \right).$$
This is the correct equation which you have to start with. Now
$$v=\frac{dx}{dt}=\pm\sqrt{v_{i}^2+\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{x_i} \right)}.$$
Assuming $v_i=0$ and integrating again from $(t=0,x_i)$ to $(t,x)$ we obtain
$$t=-\int_{x_i}^{x(t)}\frac{dx}{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{x_i} \right)}},$$
where I am using the minus sign because the axis is oriented upwards. To solve this integral you use the substitution $x=x_i\sin^2{\theta}$, 
$$t=-\sqrt{\frac{2mx_i^3}{K}}\int_{\frac{\pi}{2}}^{\theta(x)}\sin^2{\theta}d\theta,$$
where $\theta(x)=\arcsin{\sqrt{\frac{x}{x_i}}}$. Therefore,
$$t=\sqrt{\frac{mx_i^3}{2K}}\left[\frac{\pi}{2}-\arcsin{\sqrt{\frac{x}{x_i}}}-\frac 12 \sin\left(2\arcsin{\sqrt{\frac{x}{x_i}}}\right)\right].$$
However this equation cannot be solved for $x$.