From a statistical standpoint, the mean energy of a system is given by
$$\langle E \rangle = E \cdot P(x) = \frac{\int\limits_{-\infty}^{+\infty} E \mathrm e^{- \beta E}}{\int\limits_{-\infty}^{+\infty}\mathrm e^{-\beta E}},\tag{1}$$
where $\beta = 1/(k_\mathrm{B}T)$ and $P(x)$ is the probability of the system being at a particular energy $E$.
Now, if your energy dependence is quadric in some variable, this is if $E=ax^2$, where $a$ is just some constant, the mean energy becomes (thanks to some pretty cool math and Gaussian integrals)
$$\langle E \rangle = \frac{\int\limits_{-\infty}^{+\infty} {ax^2\mathrm e^{- \beta ax^2}}}{\int\limits_{-\infty}^{+\infty}\mathrm e^{- \beta ax^2}} = \frac{1}{2 \beta}= \frac{1}{2}k_\mathrm{B}T.\tag{2}$$
You should note that this result is independent of $x$ and $a$. Actually, it is easy to show that if instead of one variable, the energy depends on $n$ quadratic variables, often called the modes of the system, each mode contributes the same amount of energy $k_\mathrm{B}T/2$ to the system (all you have to do is repeat the calculations for $\left.E = \sum_{i=0}^n {a_i}x_i^2\right)$ — this is known as the equipartition theorem. Hence, the mean energy of a system with $n$ quadratic modes becomes
$$\langle E \rangle = \frac{1}{2}nk_\mathrm{B}T.\tag{3}$$
In an ideal monoatomic gas the internal energy of the system $U$ is just its kinetic energy $E_\mathrm{k}$ (there is no energy due to vibration, rotation and intermolecular interactions), and therefore
$$E=E_\mathrm{k}=\frac{1}{2}mv^2 = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2.\tag{4}$$
The energy is the sum of three quadratic modes, so the internal energy of the system is simply (substituting $n$ by $3$ in our expression for $U$)
$$U = \frac{3}{2}k_\mathrm{B}T = \frac{3}{2}nRT.\tag{5}$$
P.S. This might have been a bit more calculus then you'd asked for, but it is the reason this expression exists. I always find it helpful to know where things come from, so I hope this helps you as well.
