I will give it a try:

let \$D = \frac{R}{2L}\$ and \$\omega^2 = \frac{1}{LC}\$

$$I(s) = \frac{E}{s^2+ 2Ds + \omega^2}=$$  $$=\frac{E}{\left(s+\left(-D+\sqrt{D^2 - \omega^2}\right)\right)\left(s+\left(-D-\sqrt{D^2 - \omega^2}\right)\right)} =$$ (partial fraction) $$ =\frac{E}{-2\sqrt{D^2 - \omega^2}} \frac{1}{\left(s+\left(-D+\sqrt{D^2 - \omega^2}\right)\right)} + \frac{E}{-2\sqrt{D^2 - \omega^2}} \frac{1}{\left(s+\left(-D-\sqrt{D^2 - \omega^2}\right)\right)}$$

Looking up the Laplace transform $$ \mathcal{L}^{-1}\left[\frac{1}{s+a}\right] = e^{-at} $$

$$ \mathcal{L}^{-1}[I(s)] = \frac{E}{-2\sqrt{D^2 - \omega^2}} \left( e^{t\left(-D+\sqrt{D^2 - \omega^2}\right)} - e^{t\left(-D-\sqrt{D^2 - \omega^2}\right)} \right)$$

I will give it a try:

let \$D = \frac{R}{2L}\$ and \$\omega^2 = \frac{1}{LC}\$

for \$D^2 \not= \omega^2\$:

$$I(s) = \frac{E}{s^2+ 2Ds + \omega^2}=$$ 

$$=\frac{E}{\left(s+\left(-D+\sqrt{D^2 - \omega^2}\right)\right)\left(s+\left(-D-\sqrt{D^2 - \omega^2}\right)\right)} =$$ (partial fraction) $$ =\frac{E}{-2\sqrt{D^2 - \omega^2}} \frac{1}{\left(s+\left(-D+\sqrt{D^2 - \omega^2}\right)\right)} + \frac{E}{-2\sqrt{D^2 - \omega^2}} \frac{1}{\left(s+\left(-D-\sqrt{D^2 - \omega^2}\right)\right)}$$

Looking up the Laplace transform $$ \mathcal{L}^{-1}\left[\frac{1}{s+a}\right] = e^{-at} $$

$$ \mathcal{L}^{-1}[I(s)] = \frac{E}{-2\sqrt{D^2 - \omega^2}} \left( e^{t\left(-D+\sqrt{D^2 - \omega^2}\right)} - e^{t\left(-D-\sqrt{D^2 - \omega^2}\right)} \right)$$

for \$ D^2 = \omega^2\$:

$$I(s) = \frac{E}{s^2+ 2Ds + D^2}=\frac{E}{(s+D)^2}$$

Looking up the Laplace transform $$ \mathcal{L}^{-1}\left[\frac{1}{(s+a)^{n+1}}\right] = \frac{t^n}{n!} e^{-at} $$

$$ \mathcal{L}^{-1}[I(s)] = E \cdot t \cdot e^{-D t}$$