Inverse Laplace transform of $\frac{s}{(s + 1)^2 - 4}$

Question

I am trying to find the inverse Laplace transform $\mathcal{L}^{-1} \left\{ \dfrac{s}{(s + 1)^2 - 4} \right\}$. My textbook says that the solution is $e^{-t} \cosh(2t) - \dfrac{1}{2}e^{-t}\sinh(t)$.

But I think this is incorrect. If this is incorrect, can someone please help me find the correct one?

Thank you for any help.

Answer 1

Hint: $$\frac{s}{(s+1)^2-2^2} = \frac{s}{(s-1)(s+3)} = \frac{1}{4(s-1)}+\frac{3}{4(s+3)},$$ and $$\mathcal{L}(e^{at}) = \frac{1}{s-a}.$$

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For the textbook solution, it is useful to write the given fraction as $$\frac{s}{(s+1)^2-2^2} = \frac{s+1-1}{(s+1)^2-2^2} = \frac{s+1}{(s+1)^2-2^2} - \frac{1}{2}\cdot\frac{2}{(s+1)^2-2^2},$$ and note that $$\mathcal{L}(e^{-at}\cosh(bt)) = \frac{s+a}{(s+a)^2-b^2},$$ and $$\mathcal{L}(e^{-at}\sinh(bt)) = \frac{b}{(s+a)^2-b^2}.$$

Answer 2

$$\mathcal{L}^{-1} \left\{ \dfrac{s}{(s + 1)^2 - 4} \right\}$$

$$=\mathcal{L}^{-1} \left\{ \dfrac{s+1}{(s + 1)^2 - 4} \right\}-\mathcal{L}^{-1} \left\{ \dfrac{1}{(s + 1)^2 - 4} \right\}$$

$$=e^{-t} \cosh(2t) - \dfrac{1}{2}e^{-t}\sinh(2t)$$