The correct answer is $\boxed{{1 \over s}\left[ {F\left( s \right) – f\left( 0 \right)} \right]}$.
The Laplace transform of a function $f(t)$ is defined as
$$Lf(t): s = \int_0^\infty f(t) e^{-st} dt$$
The Laplace transform of the integral $\int_0^t f(\tau) d\tau$ is then given by
$$\begin{align} L\left \int_0^t f(\tau) d\tau \right: s &= \int_0^\infty \int_0^t f(\tau) e^{-st} d\tau dt \\ &= \int_0^\infty \frac{t}{t+s} f(t) dt \\ &= \frac{1}{s} \int_0^\infty f(t) \left( 1 – e^{-st} \right) dt \\ &= \frac{1}{s} \left[ F(s) – f(0) \right] \end{align}$$
where $F(s)$ is the Laplace transform of $f(t)$.
Option A is incorrect because it does not take into account the initial condition $f(0)$.
Option B is incorrect because it is the Laplace transform of $f(t)$, not the integral $\int_0^t f(\tau) d\tau$.
Option C is incorrect because it is the Laplace transform of the function $t f(t)$, not the integral $\int_0^t f(\tau) d\tau$.