Q:

What is the name of this circuit?

simulate this circuit – Schematic created using CircuitLab
I've tried to Google it but my keywords don't really yeild anything.
I'd like to read more about this type circuit ; what applications its useful in ? How to select C1 ?  From what I recall, it provides a DC gain of 1, but provides an AC gain set by the feedback resistors.
Does it have a name ?
Added
I should have emphasized that the focus for this question is C1 and its location in the circuit.

A:

Yes - it has a name. In control theory this circuit is known as a PD-T1 unit. It has a proportional-derivative behaviour with a certain delay term T1. In filter terms, it works like a first-order high-pass with a superimposed constant gain.
The transfer function is \$H(s)= 1 + sR1 \cdot \dfrac{C}{1+sR2C}\$
This device is used to enhance the phase (for stabilizing purposes) in a certain frequency range. Please note that application as a PD-T1 element requires \$R1>R2\$.
More than that, the shown circuit is used as a simple non-inverting amplifier (gain: \$1+R1/R2\$) for single-supply operation. For this purpose, the non-inv. input is dc biased with 50% of the supply voltage - with the consequence that the input signal must be coupled via an input capacitor. Because the dc gain remains unity, the bias voltage is transferred to the output also with the gain of "1".
BODE diagram: The magnitude starts at unity and begins to rise at \$wz=\dfrac{1}{(R1+R2)C}\$, then it stops rising at \$wp=1/R2C\$ at a gain value of \$1+(R1/R2)\$. The rising of the gain is connected with a corresponding phase enhancement.
Because of the mentioned phase enhancement properties the PD-T1 block is also known as a "lead controller".

A:

I'd just call that a non inverting amplifier.
Calculating the transfer function is quite easy if we can consider the op amp ideal.
In DC C1 is open, so you don't have current in R1 nor R2, so the op amp is in the buffer configuration and the gain is 1.
When f gets very big C1 is closed, the gain of the circuit is the usual 1+R1/R2 leading to a 2 gain for your values.
You then expect a finite pole and a finite zero, the zero comes first than the pole kicks in. You can calculate the pole with the "seen resistance" method: C1 sees R1+R2 so \$\omega_p=\frac{1}{(R_1+R_2)C_1}\$. You can now calculate the zero pulsation as \$\omega_z=\omega_p\frac{A_0}{A_\infty}=\omega_p\frac{1}{2}\$

