Find the given Laplace Transform

Question

I thought that I had almost figured this problem out. Here is my problem:

$\mathcal {L}^{-1} \left\{ \frac{1}{(s^2+1)(s^2+25)} \right\}$

The final answer that I had got was $\frac{1}{24 sin(t) -\frac{1}{120} cos(5t)}$, but was incorrect. I feel like I had maybe left something off towards the end. Any help is very much appreciated.

Answer 1

$$F(s) = \frac{1}{(s^2+1)(s^2 + 25)}$$

Unravel this using partial fractions: $$F(s) = \frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+25} = \frac{As}{s^2+1}+\frac{B}{s^2+1}+\frac{Cs}{s^2+25}+\frac{D}{s^2+25}$$

Calculate$A, B, C, D$ in that step.

Can you use the Inverse Laplace Transform tables to solve it beyond this?

Answer 2

$$\frac{1}{(s^2+1)(s^2+25)} = \frac{1}{24}\left(\frac{1}{s^2+1}-\frac{1}{s^2+25}\right) $$ and since $\mathcal{L}^{-1}\left(\frac{1}{1+s^2}\right)=\sin x$, the inverse Laplace transform of the given function is $$ \frac{1}{24}\left(\sin x-\frac{1}{5}\sin(5x)\right).$$