Circuit analysis of op amp non-inverting integrator
With the Laplace transform the capacitors have impedance

$$Z = \frac{1}{sC}$$

Use this impedance for the capacitors and analyze the circuit the same way as you would with regular resistors.

For an ideal op amp the inputs \$v_{+}\$ and \$v_{-}\$ are at the same voltage and have infinite impedance. Consequently there is a voltage divider at the input:

$$v_{+}(s) = \frac{\frac{1}{sC}}{\frac{1}{sC} + R}v_s(s) = \frac{1}{1 + sRC}v_s(s)$$

Similarly there is a voltage divider with the output:

$$v_{-}(s) = \frac{R}{R + \frac{1}{sC}}v_{o}(s) = \frac{sRC}{sRC + 1}v_{o}(s)$$

Since \$v_{+}(s) = v_{-}(s)\$ set the two equations equal to each other:

$$\frac{1}{1 + sRC}v_s(s) = \frac{sRC}{sRC + 1}v_{o}(s)$$

Re-arrange to find \$v_{o}(s)\$:

$$v_{o}(s) = \frac{1}{sRC}v_{s}(s)$$

Take the inverse Laplace transform to get back to the time domain (remember that dividing by \$s\$ is equivalent to integration):

$$v_{o}(s) = \frac{1}{sRC}v_{s}(s) \longleftrightarrow v_{o}(t) = \frac{1}{RC}\int_{0}^{t}v_{s}(x)dx + v_{o}(0)$$