Conditional density of bivariate normal and normal 
Let $Z=X+Y$ where $X \sim N(\mu,\sigma^2)$ and $Y \sim N(0,1)$ are independents. What is the
  conditional density of X given Z, $f_{X|Z}(x|z)$?

I already found that $f_{X,Z}(x,z)=\frac{1}{2\pi\sigma}e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+\sigma^2(x-z)^2]}$ and $Z \sim N(\mu,\sigma^2+1)$
$$f_{X|Z}(x|z)=\frac{f_{X,Z}(x,z)}{f_Z(z)}=\frac{\frac{1}{2\pi\sigma}e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+\sigma^2(x-z)^2]}}{\frac{1}{\sqrt{2\pi}\sqrt{\sigma^2+1}}e^{-\frac{1}{2(\sigma^2+1)}(z-u)^2}}$$
$$f_{X|Z}(x|z)=\frac{\sqrt{\sigma^2+1}}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+\sigma^2(x-z)^2]+\frac{1}{(2\sigma^2+1)}(z-\mu)^2}$$
After some simplifications I get
$$f_{X|Z}(x|z)=\frac{\sqrt{\sigma^2+1}}{\sqrt{2\pi}\sigma}e^{\frac{1}{2\sigma^2(\sigma^2+1)}[-(\sigma^2+1)(x^2-2\mu x+\sigma^2(x^2-2zx)-2\sigma^2\mu z]}$$
from here I'm stuck,Is there any easier way to find the conditional?
 A: $X$ and $Z =X+Y$ enjoy a bivariate normal density where 
$X\sim N(\mu,\sigma^2)$, $Z \sim (\mu,\sigma^2+1)$, and
$$\operatorname{cov}(X,Z)=\operatorname{cov}(X+Y,X)
= \operatorname{cov}(X,X) + \operatorname{cov}(Y,X)
= \operatorname{cov}(X,X)+0 = \sigma^2.$$
If you are allowed to apply the well-known result that
the conditional density of $X$ given $Z = z$
is a normal density with mean 
$$\mu_X + \left.\left. 
\frac{\operatorname{cov}(X,Z)}{\operatorname{var}(Z)}\right(z-\mu_Z\right)
= \mu+\left.\left.\frac{\sigma^2}{\sigma^2+1}\right(z-\mu
\right) = \frac{\sigma^2}{\sigma^2+1}z +\frac{\mu}{\sigma^2+1}\tag{1}$$
and variance
$$\operatorname{var}(X)\left(1-\frac{(\operatorname{cov}(X,Z))^2}{\operatorname{var}(X)\operatorname{var}(Z)}\right)
= \sigma^2\left(1-\frac{\sigma^4}{\sigma^2(\sigma^2+1)}\right)= \frac{\sigma^2}{\sigma^2+1}\tag{2}$$
then you can just write down the density. Just be careful with
the algebra; it is not hard, just tediously lengthy.
If you are not allowed to use this result, you can proceed as
you described in your question and just try to massage the
result into the final form that you now know it must have.
I note, for example, that the multiplicative factor is
correct in your "final answer" since
$$\frac{\sqrt{\sigma^2+1}}{\sqrt{2\pi}\sigma}
= \frac{1}{\sqrt{2\pi}\sqrt{\frac{\sigma^2}{\sigma^2+1}}}$$
is consistent with the variance given in $(2)$, but I am less
sanguine about the multiplicative factor
$\displaystyle\frac{1}{2\sigma^2(\sigma^2+1)}$ in the exponent: unless
something comes out after simplifying the rest of the exponent that
modifies the multiplicative factor,
the factor should be $\displaystyle\frac{\sigma^2+1}{2\sigma^2}$
to be consistent with $(2)$.
