(1) $L^{-1}\bigg[\frac{f(s)}{s}\bigg] = \int\limits_{0}^{t}f(t) dt$..............(1)

We know $L^{-1}[f(s)] = f(t)$, hence Also if

$L^{-1}\bigg[\frac{d^n}{ds^n}f(s)\bigg] = (-1)^n t^n f(t)$........(2)

Now $tanh^{-1}(s) = \frac{1}{2}log\bigg(\frac{1 + s}{1 - s}\bigg)$

Here f(s) = tanh^{-1}(s)

$\therefore \frac{d}{ds}f(s) = \frac{d}{ds}tanh^{-1}(s)$

$\therefore \frac{d}{ds}f(s) = \frac{d}{ds}\frac{1}{2}log\bigg(\frac{1 + s}{1 - s}\bigg)$

$\therefore \frac{d}{ds}f(s) = \frac{1}{2}\frac{d}{ds} [log(1 + s) - log(1 - s)]$

$\therefore \frac{d}{ds}f(s) = \frac{1}{2}\bigg[\frac{1}{1 + s} - \frac{1}{1 - s}\bigg]$

$\therefore \frac{d}{ds}f(s) = \frac{1}{2}\bigg[\frac{1}{s + 1} - \frac{1}{s - 1}\bigg]$

$\therefore \frac{d}{ds}f(s) = \frac{1}{2}\bigg[\frac{s - 1 - s - 1}{s^2 - 1}\bigg]$

$\therefore \frac{d}{ds}f(s) = \frac{1}{2} * \frac{-2}{s^2 - 1}$

Taking Inverse Laplace Transform,

$\therefore L^{-1}\bigg[\frac{d}{ds}f(s)\bigg] = L^{-1}\bigg[\frac{-1}{s^2 - 1}\bigg]$

$(-1)tf(t) = -sinht$ [From equation 2]

$\therefore f(t) = \frac{sinht}{t}$

As we can also write as $f(u) = \frac{1}{u}sinhu$

Now,

$L^{-1}\bigg[\frac{f(s)}{s}\bigg] = L^{-1}\bigg[\frac{tanh^{-1}(s)}{s}\bigg] = \int\limits_{0}^{t} f(u) du$

$\therefore L^{-1}\bigg[\frac{tanh^{-1}(s)}{s}\bigg] = \int\limits_{0}^{t}\frac{1}{u}sinh udu$

(2) Let us find,

$L^{-1}\bigg[\frac{s}{s^2 + 2s + 2}\bigg] = L^{-1}\bigg[\frac{s}{(s + 1)^2 + 1}\bigg]$

$\therefore L^{-1}\bigg[\frac{s}{s^2 + 2s + 2}\bigg] = L^{-1}\bigg[\frac{s + 1 - 1}{(s + 1)^2 + 1}\bigg]$

$\therefore L^{-1}\bigg[\frac{s}{s^2 + 2s + 2}\bigg] = L^{-1}\bigg[\frac{s + 1}{(s + 1)^2 + 1}\bigg] - L^{-1}\bigg[\frac{1}{(s + 1)^2 + 1}\bigg]$

$\therefore L^{-1}\bigg[\frac{s}{s^2 + 2s + 2}\bigg] = e^{-t}cost - e^{-t}sint$

$\therefore L^{-1}\bigg[\frac{s}{s^2 + 2s + 2}\bigg] = e^{-t}(cost - sint)$.......[by using first shifting property]

Now,

$L^{-1}\bigg[e^{-2s} * \frac{s}{s^2 + 2s + 2}\bigg] = e^{-(t-2)} [cos(t - 2) - sin(t - 2)]$......[using second shifting property]

$L^{-1}[e^{as} f(s)] = f(t - a) h(t - a)$.......[second shifting theorem]