I have this Laplace transform $$X(s)=\frac{1}{s\cdot(s^{2}+0.2s+1)}$$
I want to find its inverse transform. I did the following. First I decomposed that into partial fractions
$$X(s)=\frac{1}{s}-\frac{0.2+s}{s^{2}+0.2s+1}=\frac{1}{s}-\frac{0.2}{s^{2}+0.2s+1}-\frac{s}{s^{2}+0.2s+1}$$
Then I completed the square in the denominator of the second fraction
$$X(s)=\frac{1}{s}-\frac{0.2}{\sqrt{1-0.1^{2}}}\frac{\sqrt{1-0.1^{2}}}{(s+0.1)^{2}+1-0.1^{2}}-\frac{s}{(s+0.1)^{2}+1-0.1^{2}}$$
The first term is the step function in the time domain. The second one corresponds to $\frac{2\sqrt{11}}{33}\cdot e^{-\frac{t}{10}}\cdot sen\left(\frac{3\sqrt{11}}{10}t\right)$, if I'm not mistaken. However, I don't know how to inverse the third part, $\frac{s}{(s+0.1)^{2}+1-0.1^{2}}$.
How can it be done?
your answer is almost perfect just a little thing. as you said after Partial Fraction Expansion we get: $$\frac{1}{s} + \frac{-s-0.2}{s^2+0.2s+1}$$ with the knowledge of: $$\mathscr L^{-1} \{ \frac{1}{s}\}=u(t) \; $$
$$\mathscr L^{-1} \{ \frac{s}{s^2+a^2}\}=cos(at)u(t) $$ $$\mathscr L^{-1} \{ \frac{a}{s^2+a^2}\}=sin(at)u(t) $$ where $u(t)$ is unit step function or Heaviside Step Function
and also knowing that $$if \; \mathscr L\{f(t)\}=F(s) \; then \; \Rightarrow \mathscr L\{e^{bt}f(t)\}=F(s-b)$$
for example: $$\mathscr L^{-1} \{ \frac{s+b}{(s+b)^2+a^2} \}=e^{-bt}cos(at)u(t)$$ we can conclude that $$\mathscr L^{-1} \{ \frac{1}{s}+\frac{-(s+0.1)}{(s+0.1)^2+0.99}+ \frac{-(0.1)}{(s+0.1)^2+0.99}\}$$ $$= \mathscr L^{-1} \{ \frac{1}{s}+\frac{-(s+0.1)}{(s+0.1)^2+0.99}+ \frac{-(0.1)*3\sqrt{0.11}}{(s+0.1)^2+0.99}\times \frac{1}{3\sqrt{0.11}}\} $$ $$=u(t)-e^{0.1t}cos(3\sqrt{0.11}t)u(t)-\frac{0.1}{3\sqrt{0.11}}e^{0.1t}sin(3\sqrt{0.11}t)u(t)$$