What is the Inverse Laplace Transform of 1/(z^2)(z^2+1) using Residues?

[chocok]

<Inverse Laplace Transform>

Hi, I was given this: F(z) = 1/(z^2)(z^2+1) and was asked to use its residues to compute f(t). [z is a complex number]

The answer I got is -sin(t). But the answer on the book says t-sin(t).
I double checked and the residue of z=0 is 0, I don't get where the t is from.
Please, can anyone help? Thanks

 

[benorin]

$$\frac{1}{z^2(z^2+1)}=\frac{1}{z^2}-\frac{1}{z^2+1}$$

hence f(t)=t-sin(t).