Usually when performing inverse Laplace transforms, I decompose the function into partial fractions, and then look up standard transforms in a table. For example:
$$Y(s) = L^{-1}\left\{\frac{5}{s^2+3s+2}\right\} = L^{-1}\left\{\frac{5}{(s+1)(s+2)}\right\} = L^{-1}\left\{\frac{A}{s+1} + \frac{B}{s+2}\right\} $$
To find $A$ and $B$:
$$5=A(s+2)+B(s+1)$$
$$\text{let s=-2}\:\:\:\:\:\:5=B(-2+1)$$
$$B=-5$$
$$\text{from observation}\:\:\:\:\:\:A=5$$
So my original function can be written as:
$$Y(s) = 5L^{-1}\left\{\frac{1}{s+1}\right\} – 5L^{-1}\left\{\frac{1}{s+2}\right\} $$
Looking up in a table of standard transforms it can be seen:
$$y(t) = 5(e^{-1t}+e^{-2t})$$
But how do you decompose a function into partial fractions when the denominator cannot be factored?
For example the function:
$$Y(s) = L^{-1}\left\{\frac{60+12s}{s^2 + 10s + 50}\right\}$$
Wolfram alpha gives the answer as:
$$12e^{-5t}cos(5t)$$
So I need to write the function like this:
$$\frac{60+12s}{s^2 + 10s + 50} => A\frac{1}{s-a} * B\frac{s}{s^2 + b^2}$$
But how do I find $a$, $b$, $A$, and $B$?
Also, without being given the answer how am I supposed to know the form of the final answer?
So if I didn't know the form of the final answer what would be the best strategy to decompose the function?
Best Answer
If the denominator is a quadratic, you should find suitable entries in any standard Laplace table: $$L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin at\ ,\quad L^{-1}\left\{\frac{s}{s^2+a^2}\right\}=\cos at\ .$$ You can convert your example into this form by completing the square, $$L^{-1}\left\{\frac{60+12s}{s^2 + 10s + 50}\right\} =L^{-1}\left\{\frac{12(s+5)}{(s+5)^2+25}\right\}$$ and using the "shifting on the $s$-axis" rule, $$L^{-1}\bigl(F(s+a)\bigr)=e^{-at}f(t)$$ where $f(t)=L^{-1}(F(s))$.
See if you can take it from here.
Note: I have assumed that $60s+12s$ was a typo for $60+12s$.
In the last part of your question it looks as if you are assuming that $L(fg)=L(f)L(g)$. This is definitely not true!!!