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Here's how, using an approximate patched conic technique.

The $$\Delta V$$ using instantaneous (e.g. chemical propulsion) maneuvers can be determined by repeated application of this equation that simply says that the total energy is the sum of the kinetic energy and the potential energy:

$$\mathcal{E}=\frac{v^2}{2}-\frac{\mu}{r}$$

where $$\mathcal{E}$$ is the total energy per unit mass of the object or the "specific energy", $$v$$ is the velocity of the object at the current position, $$\mu$$ is the GM of the central body, i.e. Newton's gravitational constant times its mass, and $$r$$ is the current distance from the center of the central body.

The key is that the total energy of the object is a constant of motion over the orbit.

We will also use the fact that orbits are ellipses, and this equation, which determines that constant of motion from the apses of the orbit, i.e. the radii of the closest and farthest points in the orbit, $$r_1$$ and $$r_2$$:

$$\mathcal{E}=-{\mu\over r_1+r_2}$$

For an escape or inbound from escape trajectory:

$$\mathcal{E}={v_\infty^2\over 2}$$

where $$v_\infty$$ is the velocity at infinity relative to the body.

For this problem we define:

$$\mu_E$$ = GM of the Earth.
$$\mu_M$$ = GM of the Moon.
$$r_E$$ = low Earth orbit radius.
$$r_M$$ = low Mars orbit radius.
$$a_M$$ = semi-major axis (mean radius) of Moon orbit around Earth.

For simplicity, we will assume that the Moon's orbit is circular, which is not far from the truth.

Using the above, in low-Earth orbit we have for the orbital velocity $$v_{LEO}$$:

$$-{\mu_E\over 2r_E}=\frac{v_E^2}{2}-\frac{\mu_E}{r_E}$$

which gives:

$$v_{LEO}=\sqrt{\mu_E\over r_E}$$

Similarly for the low-lunar orbit velocity $$v_{LLO}$$ we get:

$$v_{LLO}=\sqrt{\mu_M\over r_M}$$

And for the velocity of the Moon in its orbit around the Earth:

$$v_M=\sqrt{\mu_E\over a_M}$$

The Hohmann transfer from the Earth to the Moon is half of an elliptical orbit about the Earth with periapsis $$r_M$$ and apoapsis $$a_M$$. For the velocity $$v$$ at any radius $$r$$ in that orbit, we have:

$$-{\mu_E\over r_E+a_M}=\frac{v^2}{2}-\frac{\mu_E}{r}$$

The velocity in that transfer orbit at the radius of a low-Earth orbit, i.e. its periapsis, is:

$$v_p=\sqrt{2a_M \mu_E\over r_E\left(a_M+r_E\right)}$$

Similarly, the velocity in that transfer orbit at the radius of the Moon, i.e. its apoapsis, is:

$$v_a=\sqrt{2r_E \mu_E\over a_M\left(a_M+r_E\right)}$$

The velocity to leave low-Earth orbit to get on the transfer orbit is:

$$\Delta V_{inject}=v_p-v_{LEO}$$

That's all you need to fly by the Moon, which is the question title. Though you ask in the question body how to get into low-Lunar orbit. You need another maneuver and more propellant to slow down and insert into an orbit.

The velocity relative to the Moon on approach, if the Moon weren't there, is:

$$v_\infty=v_M-v_a$$

That gives for the velocity on approach when the Moon is there, for any radius from the Moon:

$${v_\infty^2\over 2}=\frac{v^2}{2}-\frac{\mu_M}{r}$$

At the radius of a low-lunar orbit, that velocity is:

$$v_L=\sqrt{\left(v_M-v_a\right)^2+{2\mu_M\over r_M}}$$

To insert into orbit, we need to slow down relative to the Moon by:

$$\Delta V_{insert}=v_L-v_{LLO}$$

The total $$\Delta V$$ is then:

$$\Delta V_{total}=v_p-v_{LEO}+v_L-v_{LLO}$$

Plugging in the numbers, and assuming 200 km LEO and 100 km LLO altitudes, we get:

$$\Delta V_{inject}=3.13\,\mathrm{km\over s}$$
$$\Delta V_{insert}=0.82\,\mathrm{km\over s}$$
$$\Delta V_{total}=3.95\,\mathrm{km\over s}$$

This is close to the answer you get when doing a full integration, allowing the lunar gravity to start pulling on the spacecraft well before it gets to the distance of the Moon. That increases the speed of the spacecraft relative to the Moon a little, increasing the $$\Delta V$$ to insert a little.

This is a direct transfer that can complete in days. If you're willing to take a few months, there are lower $$\Delta V$$ paths that go through generalized Lagrange points. You can save on the order of $$0.1\,\mathrm{km\over s}$$.