Second order linear ODE equal to pulse funcion with Laplace transform method

The question is to solve the following initial conditions problem with the Laplace transform method.

$$f'' + 2f' -3f = \begin{cases} 1, \ 0 \leq t < c \\ 0, \ t \geq c \end{cases}; f(0) = f'(0) = 0$$

What I did was apply the Laplace transform to both sides so we get:

$$\mathcal{L}\{ f \}(s) \cdot (s^2 + 2s - 3) = \int_0^c e^{-st} dt = \frac{1-e^{-sc}}{s} \implies \mathcal{L}\{ f \} = \frac{1 - e^{-sc}}{s(s+3)(s-1)}$$

I have no idea how to continue from here. I wasn't able to figure out the inverse transform from the basic properties (Linearity, derivative, primitive, frequency translation, time translation, etc).

Thanks in advance for any responses.

You should decompose the rational function in partial fractions $$\frac{1}{s(s+3)(s-1)}=-\frac{1}{3s}+\frac{1}{12(s+3)}+\frac{1}{4(s-1)}$$ then consider the inverse Laplace transform $$\mathcal{L}^{-1}\left\{\frac{e^{-s\tau}}{s+s_0}\right\}=e^{-s_0(t-\tau)}H(t-\tau)$$ where $$H$$ is the unit step function.
Apply this formula for $$s_0\in\{0,3,-1\}$$ and $$\tau\in\{0,c\}.$$