Q: Inverse Laplace transform of $\frac{2-5s}{(s-6)(s^2+11)}$ What is inverse laplace transform of,
$$\frac{2-5s}{(s-6)(s^2+11)}$$
Can we split into partial fraction to solve this problem?
When I split, I get as below,
$$\frac{2-5s}{(s-6)(s^2+11)} = \frac{A}{s-6} + \frac{Bs+C}{s^2+11}$$
But, from above, does I split it correctly?

A: Yes, you can use partial fractions and yes, the ansatz you used for the partial fractions is correct.
$$\frac{2-5s}{(s-6)(s^2+11)}=\frac{A}{s-6}+\frac{Bs+C}{s^2+11}$$
$$\frac{2-5s}{(s-6)(s^2+11)}=\frac{A(s^2+11)+(Bs+C)(s-6)}{(s-6)(s^2+11)}$$
Expanding the numerator implies that:
$$\color{blue}{2}\color{green}{-5s}=\color{#990000}{As^2}+\color{blue}{11A}+\color{#990000}{Bs^2}\color{green}{-6Bs}\color{green}{+Cs}\color{blue}{-6C}$$
Comparing coefficients, we must solve the following system:
$$\begin{cases} \color{#990000}{A+B=0} \\\color{green}{-6B+C=-5} \\\color{blue}{11A-6C=2} \end{cases}$$
Solving the system is easy to do because the first equation implies $A=-B$. Doing so will give:
$$A=-\frac{28}{47} , B=\frac{28}{47} , C=-\frac{67}{47}$$
Hence:
$$\begin{align}\frac{2-5s}{(s-6)(s^2+11)}&=-\frac{28}{47}\cdot \frac{1}{s-6}+\frac{28}{47}\cdot \frac{s}{s^2+11}-\frac{67}{47}\cdot \frac{1}{s^2+11}\\ &=-\frac{28}{47}\cdot \frac{1}{s-6}+\frac{28}{47}\cdot \frac{s}{s^2+11}-\frac{67}{47\sqrt{11}}\cdot \frac{\sqrt{11}}{s^2+11} \end{align}$$

We can now apply the results 2, 7 and 8 from the Table of Laplace Transforms to give you the Inverse Laplace Transform you need. I copied the important results here:
$$\begin{array}{c|c}f(t)=\mathcal{L}^{-1}\{F(s)\}&F(s)=\mathcal{L}\{f(t)\}\\\hline e^{at}&\dfrac{1}{s-a}\\ \sin(at)&\dfrac{a}{s^2+a^2}\\ \cos(at)&\dfrac{s}{s^2+a^2} \end{array}$$
