Capacity of AWGN channel
Assuming a channel whose input at each time is a continuous random variable $X$ and its output is $Y=X+Z$, where $Z\sim\mathcal{N}(0,N)$ and $Z$ is independent of $X$, then $$C_{\text{CI-AWGN}}=\frac{1}{2}\log_2\left(1+\frac{P}{N}\right)$$ is the capacity of the continuous-input channel under the power constraint
$$\mathsf{E}X^2\le P$$
The mutual information $I(X;Y)$ is maximized (and is equal to $C_{\text{CI-AWGN}}$) when $X\sim\mathcal{N}(0,P)$.

This means that if $X$ is a continuous Gaussian random variable with the given variance, then the output has the highest possible mutual information with the input. That's it!

When the input variable $X$ is discretized (quantized), a new formulation is required. Indeed, things can easily become difficult. To see it a little bit, one can consider the simple case of a very coarse discritization of $X$ where it can only have two values. So assume that $X$ is selected from a binary alphabet, for instance let $X\in\{\pm1\}$ (or a scaled version to satisfy a power constraint). In terms of modulation, it is identical to BPSK. 

It turns out that the capacity (even in this simple case) has no closed form. I report from "Modern Coding Theory" by Richardson and Urbanke:

$$\begin{align}C&amp;_{\text{BI-AWGN}}=1+\\&amp;\frac{1}{\ln(2)}
\left(\left(\frac{2}{N}-1\right)\mathcal{Q}\left(\frac{1}{\sqrt{N}}\right)-\sqrt{\frac{2}{\pi N}}e^{-\frac{1}{2N}}+\sum_{i=1}^{\infty}\frac{(-1)^i}{i(i+1)}e^{\frac{2i(i+1)}{N}}\mathcal{Q}\left(\frac{1+2i}{\sqrt{N}}\right)\right)\end{align}$$
A comparison between the two cases can be seen in the figure below: