$$I(s)=\frac{6}{Ls^2 + Rs + \frac{1}{C}}$$
Having fixed $R$, $C$ and $L$ (for example $R=6$, $C=1/5$, $L=1$), we have:
$$I(s) = \frac{6}{s^2+6s+5}$$
First of all, find the zeros of denominator:
$$s^2 + 4s + 1= 0 \Rightarrow s = -5 \vee s = -1$$
Then $s^2+4s+1= (s+5)(s+1)$.
We can write
$$I(s) = \frac{a}{s+5} + \frac{b}{s+1} = \frac{a(s+1)+b(s+5)}{(s+1)(s+5)} = \frac{s(a+b) + (a + 5b)}{s^2 + 6s + 5} = \frac{6}{s^2+6s+5}$$
Then:
$$\left\{\begin{array}{lcl}a+b & = & 0\\a + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -b\\-b + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -\frac{3}{2}\\b & = & \frac{3}{2} \end{array}\right.$$
Then:
$$I(s) = -\frac{\frac{3}{2}}{s+5} + \frac{\frac{3}{2}}{s+1}$$
The antitransformation yield to:
$$i(t) = -\frac{3}{2}e^{-5t} + \frac{3}{2}e^{-t}$$
Best Answer
The inverse Laplace transform can be computed using the standard formula (see this page), which states that if $F$ is the Laplace transform of $f$, then $f$ can be recovered via the line integral $$f(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT} e^{tz}F(z)dz$$ where $\gamma$ is suitably chosen. In our case, $F(z)=e^{z^2/2}$. This function is smooth over the whole complex plane, so (as explained on the wikipedia page) we may take $\gamma=0$, and the line integral is along the imaginary axis from $-iT$ to $iT$. Define a parameterisation of this path in the natural way i.e. $z:[-T,T]\rightarrow\mathbb{C}$ maps $s$ to $is$. Substituting $z(s)=is$ into the line integral gives $$\int_{-iT}^{iT} e^{tz}F(z)dz=\int_{-T}^{T}e^{ist}e^{-s^2/2}z'(s)ds=i\int_{-T}^{T}e^{ist}e^{-s^2/2}ds$$ Next, write $e^{ist}=\cos(st)+i\sin(st)$. Sine is an odd function so the sine term will evaluate to zero. We are therefore left with $$f(t)=\frac{1}{2\pi i}i\int_{-\infty}^{\infty}\cos(ts)e^{-s^2/2}ds=\frac{2i}{2\pi i}\int_0^\infty \cos(\sqrt2 tx)e^{-x^2}\sqrt2dx$$ where we have used the fact that cosine is an even function, and the substitution $x=s/\sqrt2$. This integral is well known - see this question/answer. Evaluating the integral, we finally get $$f(t)=\frac{2i}{2\pi i}\sqrt2 \frac{\sqrt\pi}{2}e^{-2t^2/4}=\frac{1}{\sqrt{2\pi}}e^{-t^2/2}$$