Why do we use Laplace transforms to analyse transient circuits?
It's easy to solve simple filter circuits in the time domain but there becomes a tipping-point where most engineers would prefer to solve problems in the frequency domain and apply (for example) a step-function. Finding the inverse Laplace is fairly straightforward because of Laplace tables.

As an example of an RLC low pass filter, engineers become accustomed to the transfer function: -

$$H(s) = \dfrac{\omega_n^2}{s^2+2\zeta\omega_n^2 s + \omega_n^2}$$

And, applying (for example) a step function is as simple as multiplying by 1/s: -

$$\dfrac{1}{s}\cdot\dfrac{\omega_n^2}{s^2+2\zeta\omega_n^2 s + \omega_n^2}$$

Engineers who are familiar with this get to recognize that this converts to a standard form and the next step is just using the tables to derive the transient response.

In addition, the term \$\omega_n\$ (the natural resonant frequency) can be is factored out a lot of the time so that solving the step-function formula for \$\omega_n=1\$ begins with: -

$$\dfrac{1}{s}\cdot\dfrac{1}{s^2+2\zeta s + 1}$$

This is rearranged to a standard form such as this (underdamped resonance): -

$$\dfrac{1}{s[(s+a)^2 + b^2]}$$

Where \$a =\zeta\$ and \$b=\sqrt{1-\zeta^2}\$

The Laplace tables give us this: -

$$H(t) = 1+\dfrac{1}{\sqrt{1-\zeta^2}}\cdot e^{-\zeta t}\cdot \sin(t\cdot\sqrt{1-\zeta^2}+\phi)$$

Where \$\phi = \arccos(\zeta)\$

But you will probably only be convinced when you are dealing with slightly more complex situations or really do need to analyse the frequency spectrum.