Inverse Laplace Transform -s domain How can I find the inverse Laplace transforms of the following function?
$$ G\left(s\right)=\frac{2(s+1)}{s(s^2+s+2)}  $$
I solved so far. After that, how do I do?
$$ \frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}=G\left( s \right)$$
 A: To find the inverse Laplace transforms of the function $\ G\left(s\right)=\dfrac{2\left(s+1\right)}{s\left(s^2+s+2\right)}  $
You have solved up to partial fraction form of $G\left(s\right)$ i.e
$$G\left(s\right)=\frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}$$
Now taking the Laplace inverse $$\begin{align}\mathcal{L^{-1}}\left\{G\left(s\right)\right\}&=\mathcal{L^{-1}}\left\{\frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}\right\}\\&=\mathcal{L^{-1}}\left\{\frac{1}{s}\right\}+\mathcal{L^{-1}}\left\{\frac{1}{s^2+s+2}\right\}+\mathcal{L^{-1}}\left\{\frac{s}{s^2+s+2}\right\}\\
\end{align}$$
Now the first term $$\mathcal{L^{-1}}\left\{\frac{1}{s}\right\}=1 \\ \qquad$$ 
Second term is 
$$\begin{align}\mathcal{L^{-1}}\left\{\frac{1}{s^2+s+2}\right\}&=\mathcal{L^{-1}}\left\{\frac{1}{s^2+2\times s\times\frac{1}{2}+\left(\frac{1}{2}\right)^2+2-\left(\frac{1}{2}\right)^2}\right\}\\
&=\mathcal{L^{-1}}\left\{\frac{1}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=\mathcal{L^{-1}}\left\{\frac{2}{\sqrt{7}}\frac{\frac{\sqrt{7}}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=\frac{2}{\sqrt{7}}\mathcal{L^{-1}}\left\{\frac{\frac{\sqrt{7}}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=\frac{2}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)\\
\left[\text{since}\quad \mathcal{L^{-1}}\left\{\frac{b}{\left(s-a\right)^2+b^2}\right\}=e^{at}\sin\left(bt\right)\right]
\end{align}$$ 
Third term is
$$\begin{align}\mathcal{L^{-1}}\left\{\frac{s}{s^2+s+2}\right\}&=\mathcal{L^{-1}}\left\{\frac{s}{s^2+2.s.\frac{1}{2}+\left(\frac{1}{2}\right)^2+2-\left(\frac{1}{2}\right)^2}\right\}\\
&=\mathcal{L^{-1}}\left\{\frac{s+\frac{1}{2}-\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=\mathcal{L^{-1}}\left\{\frac{\left(s+\frac{1}{2}\right)}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}-\mathcal{L^{-1}}\left\{\frac{\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{2}\mathcal{L^{-1}}\left\{\frac{1}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{2}\frac{2}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right) \quad \left(\text{By Second term}\right)\\
&=e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)\\
\qquad \qquad \qquad \left[\text{since}\quad \mathcal{L^{-1}}\left\{\frac{\left(s-a\right)}{\left(s-a\right)^2+b^2}\right\}=e^{at}\cos\left(bt\right)\right].
\end{align}$$ Collecting all three terms, 
 $$\begin{align}\mathcal{L^{-1}}\left\{G\left(s\right)\right\}&=1+\frac{2}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)+e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)\\
&=1+\frac{1}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)+e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)\\
\end{align}$$
A: Use: 
$$
  \mathcal{LT}_s\left( \sin(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \sin(\alpha x) \mathrm{e}^{-b x} \mathrm{e}^{-s x} \mathrm{d} x = \frac{\alpha}{(s+b)^2 + \alpha^2}
$$
$$
  \mathcal{LT}_s\left( \cos(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \cos(\alpha x) \mathrm{e}^{-b x} \mathrm{e}^{-s x} \mathrm{d} x = \frac{b+s}{(s+b)^2 + \alpha^2}
$$
Completing the square: $s^2+s+2 = \left(s+\frac{1}{2}\right)^2 + \frac{7}{4}$. Therefore, decompose the image of Laplace transform accordingly:
$$
   \frac{2 (s+1)}{s
   \left(s^2+s+2\right)}=\frac{1}{s}-\frac{s+\frac{1}{2}}{\left(s+\frac{1}{2}\right)
   ^2+\frac{7}{4}}+\frac{3}{\sqrt{7}}\frac{\sqrt{7}/2}{\left(\left(s+\frac{1}{2}\right)^2+\frac{7}{4}\right)}
$$
Compare with the answer by WolframAlpha.
A: Your function $$ G(s) = \frac{2(s+1)}{s(s^2 + s + 2)} $$ has the partial fraction decomposition $$ G(s) = \frac{A}{s} + \frac{Bs + C}{(s^2 + s + 2)} $$The way how I will solve this is to use complex analysis. Your original function can be broken down into three distinct linear factors by solving for the zeros of $(s^2 + s + 2)$ using quadratic formula.
$$\frac{b\pm\sqrt{b^2 -4ac}}{2a}$$ with a = b = 1 and c = 2. The zeroes are $s = -\frac{1}{2}+j\frac{\sqrt7}{2}$ and $s = -\frac{1}{2}-j\frac{\sqrt7}{2}$. Your original function can now be written as $$ G(s) = \frac{2(s+1)}{s(s-\frac{1}{2}+j\frac{\sqrt7}{2})(s-\frac{1}{2}-j\frac{\sqrt7}{2})} $$ fixing the brackets, we have $$ G(s) = \frac{2(s+1)}{s(s-[\frac{1}{2}-j\frac{\sqrt7}{2}])(s-[\frac{1}{2}+j\frac{\sqrt7}{2}])} $$
The new partial fraction decomposition now is
$$G(s) = \frac{A}{s}+\frac{B}{[s-(\frac{1}{2}-j\frac{\sqrt7}{2})]}+\frac{C}{[s-(\frac{1}{2}+j\frac{\sqrt7}{2})]}$$ Using cover-up method, if we set $s = 0$ we will be able to solve for $A$. Thus
$$A = \frac{2(s+1)}{[s-(\frac{1}{2}-j\frac{\sqrt7}{2})][s-(\frac{1}{2}+j\frac{\sqrt7}{2})]}$$
$$A = \frac{2}{2} = 1$$ Utilize FOIL method to get this. To solve for B and C, we set $s = (\frac{1}{2}-j\frac{\sqrt7}{2}) $ and $s = (\frac{1}{2}+j\frac{\sqrt7}{2}) $ respectively. Since B and C are complex conjugates, C's imaginary part will be the negative of B. Solving for B,
$$B = \frac{2(s+1)}{s[s-(\frac{1}{2}+j\frac{\sqrt7}{2})]}$$
$$B = \frac{(1 + j\sqrt7)}{{-j\sqrt7}}$$
$$B = 1 -j\frac{\sqrt7}{7}$$
Because C is the complex conjugate of B, C is therefore
$$C = 1 +j\frac{\sqrt7}{7}$$
Thus, your original function is decomposed into
$$G(s) = \frac{1}{s} + \frac{1 - j\frac{\sqrt7}{7}}{[s-(\frac{1}{2} - j\frac{\sqrt7}{2})]} + \frac{1 + j\frac{\sqrt7}{7}}{[s-(\frac{1}{2} +j \frac{\sqrt7}{2})]}$$
The inverse Laplace Transform of this is
$$ g(t) = 1 + (1 - j\frac{\sqrt7}{7})e^{(\frac{1}{2} - j\frac{\sqrt7}{2})t} + (1 + j\frac{\sqrt7}{7})e^{(\frac{1}{2}+j\frac{\sqrt7}{2})t} $$
Do note that this is not yet the final answer. What I will be doing is I will make use of sine and cosine's equivalent complex functions, namely:
$$ sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})$$
$$ cos(\theta) = \frac{1}{2}(e^{j\theta} + e^{-j\theta})$$
We will then factor out $e^{\frac{1}{2}t}$ and distribute $e^{j\frac{\sqrt7}{2}t}$ to each terms. Upon doing so, we get
$$g(t) = 1 + e^{\frac{1}{2}t}[(1 - j\frac{\sqrt7}{7})e^{-j\frac{\sqrt7}{2}t}] + [(1 + j\frac{\sqrt7}{7})e^{j\frac{\sqrt7}{2}t}] $$
$$g(t) = 1 + e^{\frac{1}{2}t}[(e^{-j\frac{\sqrt7}{2}t} - j\frac{\sqrt7}{7}e^{-j\frac{\sqrt7}{2}t}) + (e^{j\frac{\sqrt7}{2}t} + j\frac{\sqrt7}{7}e^{j\frac{\sqrt7}{2}t}) ] $$
We will then collect the real and imaginary parts of $e^{j\frac{\sqrt7}{2}t}$ and $e^{-j\frac{\sqrt7}{2}t}$ because we will be able to get the inverse laplace transform from those. By doing so, we have
$$g(t) = 1 + e^{\frac{1}{2}t}[(e^{j\frac{\sqrt7}{2}t} + e^{-j\frac{\sqrt7}{2}t}) - j\frac{\sqrt7}{7}(e^{j\frac{\sqrt7}{2}t} - e^{-j\frac{\sqrt7}{2}t})] $$
$$g(t) = 1 + 2e^{\frac{1}{2}t}cos(\frac{\sqrt7}{2}t) - j\frac{\sqrt7}{7}(j2)e^{\frac{1}{2}t}sin(\frac{\sqrt7}{2}t) $$
$$g(t) = 1 + 2e^{\frac{1}{2}t}cos(\frac{\sqrt7}{2}t) + \frac{2\sqrt7}{7}e^{\frac{1}{2}t}sin(\frac{\sqrt7}{2}t) $$
