In engineering practice, the complex inversion integral is hardly ever used. As an engineer, you will almost exclusively need to invert rational functions, and this can be done by partial fraction expansion and elementary inversions. So first I'll show you how to obtain the inverse Laplace transform by partial fraction expansion, then I'll explain the evaluation of the inversion integral using Cauchy's residue theorem.
You have an error in the partial fraction expansion. Furthermore, you don't need to split up the complex pole pair. I would rewrite the Laplace transform like this:
$$F(s)=\frac{1}{(s+2)(s^2+4)}=\frac{A}{s+2}+\frac{Bs+C}{s^2+4}\tag{1}$$
with $A=\frac18$, $B=-\frac18$, and $C=\frac14$. The terms on the right-hand side of $(1)$ are elementary Laplace transforms. Now you just have to consider the different regions of convergence (ROC):
$$\begin{align}\frac{1}{s+2}&\Longleftrightarrow e^{-2t}u(t),&\quad \text{Re}\{s\}>-2\\\frac{1}{s+2}&\Longleftrightarrow -e^{-2t}u(-t),&\quad \text{Re}\{s\}<-2\\ \frac{s}{s^2+4}&\Longleftrightarrow \cos(2t)u(t),&\quad\text{Re}\{s\}>0\\ \frac{s}{s^2+4}&\Longleftrightarrow -\cos(2t)u(-t),&\quad\text{Re}\{s\}<0\\ \frac{1}{s^2+4}&\Longleftrightarrow \frac12\sin(2t)u(t),&\quad\text{Re}\{s\}>0\\ \frac{1}{s^2+4}&\Longleftrightarrow -\frac12\sin(2t)u(-t),&\quad\text{Re}\{s\}<0 \end{align}$$
So for the ROC $\text{Re}\{s\}<-2$ you get the anti-causal signal $$f(t)=\frac18\left[-e^{-2t}+\cos(2t)-\sin(2t)\right]u(-t)\tag{2}$$
For the ROC $-2<\text{Re}\{s\}<0$ you get the two-sided signal
$$f(t)=\frac18\left[e^{-2t}u(t)+(\cos(2t)-\sin(2t))u(-t)\right]\tag{3}$$
And, finally, for the ROC $\text{Re}\{s\}>0$ you get the causal signal $$f(t)=\frac18\left[e^{-2t}-\cos(2t)+\sin(2t)\right]u(t)\tag{4}$$
If you need to use the inversion formula then it is very helpful to know
Cauchy's residue theorem, which says that
$$\frac{1}{2\pi j}\oint_Cf(s)ds=\sum_kR_k\tag{5}$$
where $f(s)$ is analytic with finitely many poles, $C$ is a positively oriented closed curve, and $R_k$ are the residues of the poles inside $C$. It can be shown that the inversion integral equals a contour integral if the curve $C$ is chosen appropriately:
$$\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}ds= \frac{1}{2\pi j}\oint_CF(s)e^{st}ds\tag{6}$$
In the case of a rational function $F(s)$ the curve $C$ is chosen as a Bromwich contour, as shown here in Fig.2. The straight line part of the curve is the actual integration path we're interested in. The contribution from the circular part of $C$ approaches zero. Depending on the chosen ROC, we have to choose the position of the straight line (i.e., the value of $\sigma)$ differently. For ROC $\text{Re}\{s\}<-2$ (i.e., the anti-causal solution), the straight line is anywhere to the left of the left-most pole, and so the Bromwich contour enclosing all poles is negatively oriented, which results in a sign change:
$$\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}ds=-\sum_kR_k,\quad\sigma<-2,\quad t<0\tag{7}$$
where $R_k$ are the residues corresponding to the poles of the function $f(s)=F(s)e^{st}$. For ROC $\text{Re}\{s\}>0$ (i.e., the causal solution), the straight line is anywhere to the right of the right-most pole, and the Bromwich contour enclosing all poles is positively oriented. Consequently, we have
$$\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}ds=\sum_kR_k,\quad\sigma>0,\quad t>0\tag{8}$$
For the two-side solution with ROC $-2<\text{Re}\{s\}<0$ we need to choose two curves with the straight line inside the ROC. One encloses the pole to its left at $s=-2$, so the curve is positively oriented, and the other one encloses the two poles at $s=2j$ and $s=-2j$ to the right of the straight line, so it is negatively oriented (which adds a negative sign to the corresponding residues). Let $R_1$ be the residue corresponding to the pole at $s=-2$, and let $R_2$ and $R_3$ be the residues of the two poles at $\pm 2j$, respectively. The inversion integral is then given by:
$$\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}ds=\begin{cases}R_1,&t>0\\-R_2-R_3,&t<0\end{cases},\quad-2<\sigma<0\tag{9}$$
What remains is the computation of the residues. The residue at pole $p_k$ is given by $$R_k=\lim_{s\rightarrow p_k}(s-p_k)f(s)\tag{10}$$
With $f(s)=F(s)e^{st}$ we get for $p_1=-2$
$$R_1=\lim_{s\rightarrow -2}(s+2)F(s)e^{st}=\lim_{s\rightarrow -2}\frac{e^{st}}{s^2+4}=\frac{e^{-2t}}{8}\tag{11}$$
In a similar manner the other residues are obtained:
$$\begin{align}R_2&=-\frac{e^{2jt}}{8}\frac{1+j}{2}\\ R_3&=-\frac{e^{-2jt}}{8}\frac{1-j}{2}\end{align}\tag{12}$$
Now $(7)-(9)$ can be evaluated, and the results are of course the same as $(2)-(4)$.
