Inverse Laplacetransform of rational function with multiple pole I have to calculate the inverse Laplacetransorm of this function using Residue calculus
$$
\frac{s^4 + 6s^3 - 10s^2 + 1}{s^5}
$$
but I can't find any Residue rule that would solve this. Can you show me how to solve this without using partial fractions(I doubt that its even possible to do it using partial fractions).
I have used some very poorly described method from my book, that i never fully understood and never got the right results.
 A: If $F(s)=\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)$ is the Laplace Transform of $f(t)$, then the Inverse Laplace Transform is given by 
$$f(t)=\frac{1}{2\pi i}\int_{\sigma -i\infty}^{\sigma +i\infty}e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)\,ds \tag 1$$
where here $\sigma >0$. 
For $t>0$, we will close the contour with an "infinite semi-circle" that enclosed the left-half plane, while for $t<0$ we will close the contour with an "infinite semi-circle" that encloses the right-half plane.  Then, the integral $(1)$ is equal to $2\pi i$ times the residue of the integrand.  
For $t>0$, we close the contour in the left-half plane for which $F(s)$ has a pole at $s=0$.  We can find the residue by first expanding $e^{st}$ in its Taylor series
$$e^{st}=\sum_{n=0}^{\infty}\frac{(st)^n}{n!}=1+(st)+\frac12(st)^2+\frac16(st)^3+\frac{1}{24}(st)^4+\cdots \tag 2$$
and computing the coefficient on the term $s^{-1}$ of $e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)$.  Proceeding, we find that  
$$\text{Res}\left(e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right),s=0\right)=1+6t-5t^2+\frac{1}{24}t^4$$
Therefore, for $t>0$
$$f(t)=1+6t-5t^2+\frac{1}{24}t^4$$
For $t<0$, we close the contour in the right-half plane for which there are no singularities and find that $f(t)=0$ for $t<0$.
Putting it all together, we have 
$$\bbox[5px,border:2px solid #C0A000]{f(t)=\left(1+6t-5t^2+\frac{1}{24}t^4\right)u(t)}$$
where $u(t)$ is the unit-step function.

NOTE:
In this note, we address a specific request regarding the calculation of the residue.  To do this we rely on the expansion given by $(2)$.
Now, the residue is equal to the coefficient on the $\frac1s$ term of the product
$$\begin{align}
e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)&=\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)\sum_{n=0}^{\infty}\frac{(st)^n}{n!}\\\\
&=\frac1s\sum_{n=0}^{\infty}\frac{(st)^n}{n!} \tag{A}\\\\
&+\frac{6}{s^2}\sum_{n=0}^{\infty}\frac{(st)^n}{n!} \tag{B}\\\\
&-\frac{10}{s^3}\sum_{n=0}^{\infty}\frac{(st)^n}{n!}\tag{C}\\\\
&+\frac{1}{s^5}\sum_{n=0}^{\infty}\frac{(st)^n}{n!} \tag{D}
\end{align}$$
In $(A)$ the coefficient on the $s^{-1}$ term is $1$.  This result comes from multiplying $\frac1s$ by the first term in the series, $\frac{(st)^0}{0!}=1$, where $0!$ is defined to be $1$.  Thus, $$\frac1s \times \frac{(st)^0}{0!}=\frac1s$$
In $(B)$ the coefficient on the $s^{-1}$ term is $6t$.  This result comes from multiplying $\frac{6}{s^2}$ by the second term in the series, $\frac{(st)^1}{1!}=st$.  Thus, $$\frac{6}{s^2}\times \frac{(st)^1}{1!}=\frac{6t}{s}$$
In $(C)$ the coefficient on the $s^{-1}$ term is $-5t^2$.  This result comes from multiplying $-\frac{10}{s^3}$ by the third term in the series, $\frac{(st)^2}{2!}=\frac12(st)^2$.  Thus, $$-\frac{10}{s^3}\times \frac{(st)^2}{2!}=\frac{-5t^2}{s}$$
In $(D)$ the coefficient on the $s^{-1}$ term is $\frac{1}{24}t^4$.  This result comes from multiplying $\frac{1}{s^5}$ by the fifth term in the series, $\frac{(st)^4}{4!}=\frac{1}{24}(st)^4$.  Thus, $$\frac{1}{s^5}\times \frac{(st)^4}{4!}=\frac{\frac{1}{24}t^4}{s}$$
Adding the four contributions gives the expected result.
