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

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I am a math student at freshman university level and I am taking an ODE class. I am not expected to integrate in the complex plane or to know convolution (so no duplication with Inverse Laplace of $\frac 1 {(s^2+a^2)^n}$). The way we've done inverse Laplace in this class is by using the basic properties of Laplace Transform and referring to known Laplace transforms like sin, cos, and polynomials)

I don't know where to start but I know a first step is to consider that $$\mathcal{L} (t\times \cos t)= -\frac{d}{ds}\frac{s}{s^2+1}$$

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Hint

You have that $$\mathcal{L} (t\times \cos t)= -\frac{d}{ds}\frac{s}{s^2+1}=\frac {s^2-1}{(s^2+1)^2}$$ $$\mathcal{L} (\sin t)= \frac{1}{s^2+1}=\frac {s^2+1}{(s^2+1)^2}$$

Substract ...