Partial Fractions: Expanding & Laplace Transforms

[mxpxer7]

I'm sure this is a no brainer to someone, but here it is..

what is does the partial fraction of this look like in expanded form? Or how can I make it fit on the table of laplace transforms?

__(2s+1)__
(s-1)^2 + 1

 

[RedX]

I think you'd get more replies (and better ones) if you posted this in the algebra forum, as this doesn't really have anything to do with differential equations. Your case would fall under the "irreducible quadratic factor in the denominator", as the denominator has zeroes s=1+i and s=1-i which aren't real:

http://en.wikipedia.org/wiki/Partial_fraction#An_irreducible_quadratic_factor_in_the_denominator

So what you wrote is the simplest form if you want to use only real numbers.

 

[mxpxer7]

I apologize for putting this in the wrong section I'm using the answer for la place transforms so maybe this is where the calculus comes in, How can i write this so it fits into the table of laplace transforms?

 

[RedX]

$$F(s)=\frac{2s+1}{(s-1)^2+1}=2\frac{s-1}{(s-1)^2+1}+3\frac{1}{(s-1)^2+1}=G(s-1)$$

where $$G(s)=2\frac{s}{(s)^2+1}+3\frac{1}{(s)^2+1}$$

The inverse Laplace transform of G(s) is 2*cos(x)+3*sin(x). The inverse Laplace transform of G(s-1) is $$e^x$$[2*cos(x)+3*sin(x)] according to some of the properties of Laplace transforms and a shifted arguments.

 

[trambolin]

$$F(s)=\frac{2s+1}{(s-1)^2+1}=2\frac{s-1}{(s-1)^2+1}+3\frac{1}{(s-1)^2+1}=G(s-1)$$

where $$G(s)=2\frac{s}{(s)^2+1}+3\frac{1}{(s)^2+1}$$

The inverse Laplace transform of G(s) is 2*cos(x)+3*sin(x). The inverse Laplace transform of G(s-1) is $$e^x$$[2*cos(x)+3*sin(x)] according to some of the properties of Laplace transforms and a shifted arguments.

Nice shortcut...