Find a probability density I am going through a paper trying to understand all the single steps, but I got stuck. I need to calculate
$$p(x+\delta t) \mid x(t), t)= \int p(x(t+\delta t) \mid \mu , x(t), t)p(\mu\mid  x(t), t) d\mu  $$ 
where the first term is $$N(x(t+\delta t) \mid x+\mu\delta t, \delta t) $$
and the second term is $$N(\mu \mid x(t)/(t+\sigma ^{-2} ), 1/(t+\sigma ^{-2})) $$
I know that the resulting distribution is $$N(x(t+\delta t) \mid x(t)(1+\delta t /(t+\sigma ^{-2}), \delta t(1+ \delta t/(1+\sigma ^{-2}))) $$
However, I would like to know how to get there.
FYI, the paper is this one: http://www.jneurosci.org/content/32/11/3612.full and the calculation are in the paragraph "belief transition densities", by putting together pieces from previous equations. Note that there is a typo for $\delta t_{eff} $, which is equal to $\delta t /(t+ \sigma ^{-2} ) $.
 A: Writing out the Gaussian densities explicitly leads to
$$p(x+\delta t) \mid x(t), t)\\= \int \frac{1}{2\pi}\sqrt{\frac{t+\sigma^{-2}}{(\delta t)}}\exp\left(-\frac{1}{2(\delta t)}\left[x(t+\delta t)-x(t))-\mu (\delta t)\right]^2\right)\\\times\exp\left(-\frac{(t+\sigma^{-2})}{2}\left(\mu-\frac{x(t)}{t+\sigma^{-2}}\right)^2\right) d\mu$$ 
Thus
$$p(x+\delta t) \mid x(t), t)\propto \int\exp\left(-\frac{1}{2(\delta t)}\left[x(t+\delta t)-x(t))-\mu (\delta t)\right]^2\right)\\\times\exp\left(-\frac{(t+\sigma^{-2})}{2}\left(\mu-\frac{x(t)}{t+\sigma^{-2}}\right)^2\right) d\mu\text{ (Eq.1)}$$
As we wish to integrate w.r.t. $\mu$, we group the $\mu$ terms together, and complete the square (by adding and subtracting the terms in red below):-
$$\begin{align}\int&\exp\left(-\frac{\delta t+(t+\sigma^{-2})}{2}\left[\mu^2-\frac{2x(t+\delta t)}{\delta t+(t+\sigma^{-2})}\mu\color{red}{+\frac{x^2(t+\delta t)}{(\delta t+(t+\sigma^{-2}))^2}-\frac{x^2(t+\delta t)}{(\delta t+(t+\sigma^{-2}))^2}}\right]\right)d \mu\\&=\exp\left(+\frac{1}{2}\frac{x^2(t+\delta t)}{\delta t+(t+\sigma^{-2})}\right)\color{blue}{\int\exp\left(-\frac{\delta t+(t+\sigma^{-2})}{2}\left[\mu-\frac{x(t+\delta t)}{\delta t+(t+\sigma^{-2})}\right]^2\right)d \mu}\\&\propto \exp\left(+\frac{1}{2}\frac{x^2(t+\delta t)}{\delta t+(t+\sigma^{-2})}\right)\color{blue}{\int N\left(\mu\Big|\frac{x(t+\delta t)}{\delta t+(t+\sigma^{-2})},\frac{1}{\delta t+(t+\sigma^{-2})}\right) d \mu}\\&=\color{green}{\exp\left(+\frac{1}{2}\frac{x^2(t+\delta t)}{\delta t+(t+\sigma^{-2})}\right)}\end{align}$$
Having integrated out $\mu$, we substitute the above expression into (Eq.$1$), and re-use the completing the square technique to obtain the desired result
$$\begin{align}p(x+\delta t) \mid x(t), t)&\propto\exp\left(-\frac{1}{2}\left[x^2(t+\delta t)\left(\frac{1}{\delta t}-\color{green}{\frac{1}{\delta t+(t+\sigma^{-2})}}\right)-\frac{2x(t)x(t+\delta t)}{\delta t}\right]\right)\\&=\exp\left(-\frac{(t+\sigma^{-2})}{2(\delta t)(\delta t+(t+\sigma^{-2})}\left[x^2(t+\delta t)-\frac{2x(t)(\delta t+(t+\sigma^{-2}))x(t+\delta t)}{(t+\sigma^{-2})}\\\color{red}{+\frac{x^2(t)(\delta t+((t+\sigma^{-2}))^2}{(t+\sigma^{-2})^2}-\frac{x^2(t)(\delta t+((t+\sigma^{-2}))^2}{(t+\sigma^{-2})^2}}\right]\right)\\&\propto\exp\left(-\frac{(t+\sigma^{-2})}{2(\delta t)(\delta t+(t+\sigma^{-2}))}\left[x(t+\delta t)-\frac{x(t)(\delta t+(t+\sigma^{-2}))}{(t+\sigma^{-2})}\right]^2\right)\\&\propto 
N\left(x(t+\delta t)\Big|\frac{x(t)(t+\sigma^{-2})}{\delta t+(t+\sigma^{-2})},\frac{\delta t(\delta t+(t+\sigma^{-2}))}{(t+\sigma^{-2})}\right)\\&=N\left(x(t+\delta t)\Big|\frac{x(t)(\delta t+(t+\sigma^{-2}))}{(t+\sigma^{-2})},\frac{\delta t(\delta t+(t+\sigma^{-2}))}{(t+\sigma^{-2})}\right)
\\&=N\left(x(t+\delta t)\Big|x(t)\left(1+\frac{\delta t}{t+\sigma^{-2}}\right),\delta t\left(1+\frac{\delta t}{t+\sigma^{-2}}\right)\right)\end{align}$$
