Why does the inverse Laplace transform of 5/((2x+3)(4+x²)) give trig terms?

[Ry122]

Homework Statement

I'm attempting to find the inverse laplace transform of (5/((2x+3)(4+x^2)))

The Attempt at a Solution

Here's the solution:
http://www.wolframalpha.com/input/?i=inverse+laplace+transform+(5/((2x+3)(4+x^2)))

There's 3 terms in the solution and 2 are trigonometric. But according to laplace tables you
only get a trigonometric term if a variable in one of the terms of the denominator is squared and if you perform PFE there won't be any variables squared so I'm not sure what's going on here.

 

[susskind_leon]

http://www.wolframalpha.com/input/?i=%285%2F%28%282x%2B3%29%284%2Bx^2%29%29%29
The PFE does have x^2 in the denominator

 

[vela]

Did you actually do the partial fractions expansion?

 

[Ry122]

http://www.wolframalpha.com/input/?i=%285%2F%28%282x%2B3%29%284%2Bx^2%29%29%29
The PFE does have x^2 in the denominator

How do you work out that term for the PFE?
Using this method I got the correct answer for the 2nd term but not the term with x^2 as the denominator.

4+x^2 = 0
so the limit is at x = 2

sub x = 2 into
5/(2x+3) to get the value of the residual then
the final answer is 5/7(4+x^2)

 

[Ry122]

Worked it out, I needed Bx + C as the residual and not just B.