Remember, ideal op amps follow two basic rules:

1. No current flows into either input.

2. Negative feedback forces the voltage at each input to be equal.

Let's start with a qualitative approach. Since there's one capacitor, we can divide the frequency response of this circuit into three regions -- low-frequency \$(Z_C \gg R_3)\$, mid-frequency \$(Z_C \approx R_3)\$, and high-frequency \$(Z_C \ll R_3)\$.

At DC, the capacitor acts like an open circuit, so the non-inverting input is tied to ground. In this case, the negative feedback is a simple inverting amplifier.

At high frequency, the capacitor acts like a short circuit, so the non-inverting input is tied directly to \$V_{in}\$. It's harder to see, but in this case the negative feedback gives you a voltage follower.

At mid-frequency, the frequency response transitions from inverting input to voltage follower. We expect the gain to go from R2/R1 to 1 and the phase to go from 180 to 0. This is where we have to derive the transfer function using the op amp rules. \$V_+\$ is pretty easy -- C and R3 form a low-pass filter:

$$V_+ = V_{in}\frac{R_3}{R_3 + \frac{1}{sC}} = V_{in}\frac{1}{1 + \frac{1}{sR_3C}}$$

\$V_-\$ is a little trickier, but it's mostly the same as deriving an inverting amplifier:

$$\frac{V_{in} - V_-}{R_1} = \frac{V_- - V_{out}}{R_2}$$ $$(R_1 + R_2)V_- = R_1V_{out} + R_2V_{in}$$

Now we connect our two equations with:

$$V_+ = V_-$$

$$V_{in}\frac{1}{1 + \frac{1}{sR_3C}}(R_1 + R_2) = R_1V_{out} + R_2V_{in}$$

From here, it's just a matter of algebra. It's up to you how you want to express the result, but a convenient thing to end up with is:

$$\frac{V_{out}}{V_{in}} = (DC\ gain) + (AC\ gain)*(frequency\ response)$$

Here's my derivation for that:

$$V_{in}\frac{R_1 + R_2}{1 + \frac{1}{sR_3C}} = R_1V_{out} + R_2V_{in}$$

$$V_{out} = -\frac{R_2}{R_1}V_{in} + \frac{V_{in}}{R_1}\frac{R_1 + R_2}{1 + \frac{1}{sR_3C}}$$

$$V_{out} = -\frac{R_2}{R_1}V_{in} + V_{in}\frac{1 + \frac{R_2}{R_1}}{1 + \frac{1}{sR_3R_1C}}$$

$$\frac{V_{out}}{V_{in}} = -\frac{R_2}{R_1} + (\frac{R_2}{R_1} + 1)\frac{1}{1 + \frac{1}{sR_3C}}$$

When s -> 0, the gain becomes:

$$\frac{V_{out}}{V_{in}} = -\frac{R_2}{R_1} + (\frac{R_2}{R_1} + 1)*0 = -\frac{R_2}{R_1}$$

When s -> infinity, the gain becomes:

$$\frac{V_{out}}{V_{in}} = -\frac{R_2}{R_1} + (\frac{R_2}{R_1} + 1)*1 = 1$$

That's the behavior we expected to begin with, so probably I did the algebra correctly. :-)

Remember, ideal op amps follow two basic rules:

1. No current flows into either input.

2. Negative feedback forces the voltage at each input to be equal.

Let's start with a qualitative approach. Since there's one capacitor, we can divide the frequency response of this circuit into three regions -- low-frequency \$(Z_C \gg R_3)\$, mid-frequency \$(Z_C \approx R_3)\$, and high-frequency \$(Z_C \ll R_3)\$.

At DC, the capacitor acts like an open circuit, so the non-inverting input is tied to ground. In this case, the negative feedback is a simple inverting amplifier.

At high frequency, the capacitor acts like a short circuit, so the non-inverting input is tied directly to \$V_{in}\$. It's harder to see, but in this case the negative feedback gives you a voltage follower.

At mid-frequency, the frequency response transitions from inverting input to voltage follower. We expect the gain to go from R2/R1 to 1 and the phase to go from 180 to 0. This is where we have to derive the transfer function using the op amp rules. \$V_+\$ is pretty easy -- C and R3 form a low-pass filter:

$$V_+ = V_{in}\frac{R_3}{R_3 + \frac{1}{sC}} = V_{in}\frac{1}{1 + \frac{1}{sR_3C}}$$

\$V_-\$ is a little trickier, but it's mostly the same as deriving an inverting amplifier:

$$\frac{V_{in} - V_-}{R_1} = \frac{V_- - V_{out}}{R_2}$$ $$(R_1 + R_2)V_- = R_1V_{out} + R_2V_{in}$$

Now we connect our two equations with:

$$V_+ = V_-$$

$$V_{in}\frac{1}{1 + \frac{1}{sR_3C}}(R_1 + R_2) = R_1V_{out} + R_2V_{in}$$

From here, it's just a matter of algebra. It's up to you how you want to express the result, but one way that's easy to understand is:

$$\frac{V_{out}}{V_{in}} = (DC\ gain) + (AC\ gain)*(frequency\ response)$$

(Note that Lorenzo's form is probably more common in signal processing, but I like this one for educational purposes.) Here's my derivation:

$$V_{in}\frac{R_1 + R_2}{1 + \frac{1}{sR_3C}} = R_1V_{out} + R_2V_{in}$$

$$V_{out} = -\frac{R_2}{R_1}V_{in} + \frac{V_{in}}{R_1}\frac{R_1 + R_2}{1 + \frac{1}{sR_3C}}$$

$$V_{out} = -\frac{R_2}{R_1}V_{in} + V_{in}\frac{1 + \frac{R_2}{R_1}}{1 + \frac{1}{sR_3C}}$$

$$\frac{V_{out}}{V_{in}} = -\frac{R_2}{R_1} + (\frac{R_2}{R_1} + 1)\frac{1}{1 + \frac{1}{sR_3C}}$$

When s -> 0, the gain becomes:

$$\frac{V_{out}}{V_{in}} = -\frac{R_2}{R_1} + (\frac{R_2}{R_1} + 1)*0 = -\frac{R_2}{R_1}$$

When s -> infinity, the gain becomes:

$$\frac{V_{out}}{V_{in}} = -\frac{R_2}{R_1} + (\frac{R_2}{R_1} + 1)*1 = 1$$

That's the behavior we expected to begin with, which is a good sign that I did the algebra correctly. :-) You can also check with Excel or some other tool vs. what you get in CircuitLab. Doing a frequency response simulation in CircuitLab is probably the easiest way to get started with an unfamiliar filter circuit.