Sketch: Let $X=Y+Z$ with $Y,Z$ normal and independent: $Y \sim N(0,1) $ and $Z \sim N(0,a) $
We want $E[Y|X]$.
A common mistake (that you did) is to write $E[Y|X]=E[X|X]-E[Z|X]=X$ . The problem is that $X$ and $Z$ are not independent, hence $E[Z|X] \ne 0$ [*]
One way is to apply the inversion formula to the densities. Informally:
$$P(Y|X) = \frac{P(X|Y)P(Y)}{P(X)}=\frac{\psi(X;Y,a) \,\psi(Y;0,1)}{\psi(X;0,a+1)}$$
where $\psi(t;u,s)$ is the normal density function with mean $u$ and variance $s$ evaluated at $t$.
Then, assuming $X$ is fixed, we can ignore the denominator (it's a normalization factor) and the numerator is the product of normals; working out the exponent (remember that $Y$ is our variable here) when can complete the square...
$$-\frac{(X-Y)^2}{2a}-\frac{Y^2}{2}=-\frac{1}{2a}\left((1+a)Y^2-2XY+X^2\right)=\\
=-\frac{1+a}{2a}\left( Y- \frac{X}{1+a} \right)^2 + c$$
with some $c$ does not depend on $Y$ (no need to compute it).
Hence, we get that $Y$ conditioned on $X$ is normal with mean $\frac{X}{1+a}$ (you also can see the variance), so $$E[Y|X]=\frac{X}{1+a}$$
[*] Why this mistake is common? Perhaps because of a wrong use of cause-effect concepts. Given $C=A+B$ one regards $C$ as an "ouput" and $A,B$ as "inputs" (ok), then slip on to consider of $A,B$ as "causes" and $C$ as the "effect", then one considers that "the output (effect) depends on the input (cause)", but on the contrary, "the input does not depend on the output" (false; for one thing, independence is a symmetric relation). Hence to condition on the output does not changes the probabilities of the inputs, and $E(A|C)=E(A)$. Of course, this is wrong.
Best Answer
The conditional distribution $f(x|x>a)$ is given by
$$f_{X|X>a}(x)=\frac{1}{1-F_X(a)}f_X(x)\mathbb{1}_{(a;\infty)}(x)$$
The integral involved to calculate the expectation can be easy solved as it is in the form
$$\int x e^{-\frac{x^2}{2}}dx$$
and as you note, $x$ is something like the exponent derivative...
Example:
$X\sim N(1;1)$
Calculate $\mathbb{E}(X|X>1)$
$$\mathbb{E}(X|X>1)=\frac{1}{1-F(1)}\int_1^{\infty}\frac{1}{\sqrt{2\pi}}x e^{-\frac{(x-1)^2}{2}}dx=$$
Set $x-1=y$
$$2\Bigg[\int_0^{\infty}\frac{1}{\sqrt{2\pi}}ye^{-\frac{y^2}{2}}dy+\underbrace{\int_0^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dy}_{=\frac{1}{2}}\Bigg]$$
$$=2\Bigg\{-\frac{1}{\sqrt{2\pi}}\Bigg[e^{-\frac{y^2}{2}}\Bigg]_0^{\infty}+\frac{1}{2}\Bigg\}=\frac{2}{\sqrt{2\pi}}+1\approx1.797885$$
General example
$X\sim N(\mu;\sigma^2)$; Calculate $\mathbb{E}(X|X>a)$
$$\mathbb{E}(X|X>a)=\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\int_a^{\infty}\frac{1}{\sigma\sqrt{2\pi}}x e^{-\frac{1}{2}\Big(\frac{x-\mu}{\sigma}\Big)^2}$$
Set $\frac{x-\mu}{\sigma}=y$
$$\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\int_{\frac{a-\mu}{\sigma}}^{\infty}\frac{1}{\sqrt{2\pi}}(\sigma y+\mu) e^{-\frac{1}{2}y^2}dy$$
$$\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\Bigg\{\int_{\frac{a-\mu}{\sigma}}^{\infty}\frac{\sigma y}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}dy+\mu\int_{\frac{a-\mu}{\sigma}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2}dy\Bigg\}$$
$$\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\Bigg\{\frac{\sigma}{\sqrt{2\pi}}\Big[ -e^{-\frac{1}{2}y^2}\Big]_{\frac{a-\mu}{\sigma}}^{\infty}+\mu\Big[1-\Phi\Big(\frac{a-\mu}{\sigma}\Big] \Bigg\}$$
$$ \bbox[5px,border:2px solid red] { \mathbb{E}(X|X>a)=\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\Bigg\{\frac{\sigma}{\sqrt{2\pi}}e^{-\frac{1}{2}\Big(\frac{a-\mu}{\sigma}\Big)^2}+\mu\Big[1-\Phi\Big(\frac{a-\mu}{\sigma}\Big] \Bigg\} \ } $$