I'm studying circuit analysis. I've to solve this inverse Laplace transform to see the response:
$$\mathcal{L}_{s}^{-1}\left[\frac{r}{r+\frac{1}{cs}}\cdot\frac{k\tanh\left(\frac{as}{2}\right)}{s}\right]_{(t)}$$
Where $k,r,c,a\in\mathbb{R}^+$
My work:
$$\mathcal{L}_{s}^{-1}\left[k\left(\frac{1}{s+\frac{1}{cr}}\right)\left(1-\frac{2}{e^{as}+1}\right)\right]_{(t)}=e^{-\frac{t}{cr}}-2cr\mathcal{L}_{s}^{-1}\left[\frac{1}{\left(1+e^{as}\right)\left(1+crs\right)}\right]_{(t)}$$
Consider one method from where the proposer's work ends:
Starting with $$\mathcal{L}_{s}^{-1}\left[k\left(\frac{1}{s+\frac{1}{cr}}\right)\left(1-\frac{2}{e^{as}+1}\right)\right]_{(t)} = k\,e^{-\frac{t}{cr}} - 2ckr \, \mathcal{L}_{s}^{-1}\left[\frac{1}{\left(1+e^{as}\right)\left(1+crs\right)}\right]_{(t)}$$ then \begin{align} \mathcal{L}_{s}^{-1}\left[\frac{1}{\left(1+e^{as}\right)\left(1+crs\right)}\right]_{(t)} &= \sum_{j=0}^{\infty} (-1)^{j} \, \mathcal{L}_{s}^{-1}\left[\frac{e^{-a (j+1) s}}{1+crs}\right]_{(t)} \\ &= \frac{1}{c r} \, \sum_{j=0}^{\infty} (-1)^{j} \, H(t - (j+1) a) \, e^{- (t - a (j+1))/c r} \end{align} which leads to $$\mathcal{L}_{s}^{-1}\left[k\left(\frac{1}{s+\frac{1}{cr}}\right)\left(1-\frac{2}{e^{as}+1}\right)\right]_{(t)} = k\,e^{-\frac{t}{cr}} - 2k \, \sum_{j=0}^{\infty} (-1)^{j} \, H(t - (j+1) a) \, e^{- (t - a (j+1))/c r},$$ where $H(t-a)$ is the Heaviside Step function.