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

Question

While solving Volterra integral equation $$y(x)=\sin x - \int_0^x (x-t)y(t)\ \mathsf dt$$ with Laplace transformation, I found $$Y(s)= \frac{s^2}{(s^2+1)^2},$$ and I dont know to calculate its inverse Laplace transform.

Answer 1

With partial fractions you get $$\frac{1}{s^2+1}-\frac{1}{\left(s^2+1\right)^2}$$ and then looking on a table the result $$y=\frac{1}{2} (\sin t+t \cos t)$$