Which one of the options given is the inverse Laplace transform of 1/ (𝑠³−𝑠 )? 𝑢(𝑡) denotes the unit-step function.

The correct answer is:
$$(-1+\frac{1}{2}{e}^{-t}+\frac{1}{2}{e}^t)u\left(t\right)$$

Step-by-step Explanation:

We are given the Laplace transform function:
$$F\left(s\right)=\frac{1}{s^3-s}$$

First, factor the denominator completely:
$$s^3-s=s(s^2-1)=s(s-1)(s+1)$$

Next, express the function using partial fraction decomposition:
$$\frac{1}{s(s-1)(s+1)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s+1}$$

We can solve for the coefficients $A$, $B$, and $C$ using the residue method:

To find $A$, multiply the equation by $s$ and evaluate at $s=0$:
$$A={[\frac{1}{(s-1)(s+1)}}_{}]s=0=\frac{1}{(-1)\left(1\right)}=-1$$

To find $B$, multiply the equation by $s-1$ and evaluate at $s=1$:
$$B={[\frac{1}{s(s+1)}}_{}]s=1=\frac{1}{1(1+1)}=\frac{1}{2}$$

To find $C$, multiply the equation by $s+1$ and evaluate at $s=-1$:
$$C={[\frac{1}{s(s-1)}}_{}]s=-1=\frac{1}{-1(-1-1)}=\frac{1}{2}$$

Substituting the values of $A$, $B$, and $C$ back into the decomposition yields:
$$F\left(s\right)=-\frac{1}{s}+\frac{1}{2(s-1)}+\frac{1}{2(s+1)}$$

Now, apply the inverse Laplace transform to each term using the standard transform pairs:
$${\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}=u\left(t\right)$$

and
$${\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}u\left(t\right)$$

This gives:
$$f\left(t\right)=-u\left(t\right)+\frac{1}{2}{e}^{t}u\left(t\right)+\frac{1}{2}{e}^{-t}u\left(t\right)$$

Factoring out the unit-step function $u\left(t\right)$ results in the final solution:
$$f\left(t\right)=(-1+\frac{1}{2}{e}^{-t}+\frac{1}{2}{e}^t)u\left(t\right)$$