$X$ ~ $N(u, \sigma^2)$. Find $E(X)$
We know that $f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-u}{\sigma} \right)^2}$
Also,
$Z = \frac{X-u}{\sigma}$ ~ $N(0, 1)$, so $f_Z(z) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}$
$E(Z) = E\left[\frac{X}{\sigma} - \frac{u}{\sigma} \right] = \frac{1}{\sigma}E(X) + \frac{u}{\sigma}E(1) = E(X) = u$
Therefore $E(X) = u$ for $X$ ~ $N(u, \sigma^2)$
I don't get it. The part where "$\frac{1}{\sigma}E(X) + \frac{u}{\sigma}E(1) = E(X) = u$"
I'm aware that $X$ ~ $N(0, 1)$, $E(X) = 0$.
You have a very big mistake. First of all, the expectation operator is linear, therefore $$E(Z) = \frac{1}{\sigma}E(X) - \frac{\mu}{\sigma}E(1) = \frac{\mu}{\sigma}-\frac{\mu}{\sigma} =0 $$
Second of all, the title's content is different from what you are asking. So from hereon, I will address the question in the title: $$E(X) = \int\limits_{-\infty}^{+\infty} x f_X(x)$$ where $$f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2\sigma^2}(x-\mu)^2}$$ So $$E(X) = \frac{1}{\sigma\sqrt{2\pi}} \int\limits_{-\infty}^{+\infty} x e^{-\frac{1}{2\sigma^2}(x-\mu)^2} \ dx$$ Take a change of variable $t = x-\mu$ so $dt = dx$ and then $$E(X) = \frac{1}{\sigma\sqrt{2\pi}} \int\limits_{-\infty}^{+\infty} (t+\mu) e^{-\frac{1}{2\sigma^2}t^2} \ dt = \frac{1}{\sigma\sqrt{2\pi}} \int\limits_{-\infty}^{+\infty} t e^{-\frac{1}{2\sigma^2}t^2} \ dt+\mu \int\limits_{-\infty}^{+\infty} \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2\sigma^2}t^2} \ dt$$ So the first integral evaluates to zero because it is an odd function. The second integral is the integration of the Normal PDF hence it evaluates to 1. Finally $$E(X) = 0 + \mu(1) = \mu$$