Calculate Inverse Laplace Transform $H(s)=\frac{1}{s^4-s^2}$

Question

Calculate Inverse Laplace Transform $$H(s)=\frac{1}{s^4-s^2}$$

I am trying to solve this by convolution but I don't know how to move forward. I have come this far.

$$H(s) = \frac{1}{s^4-s^2}$$ $$h(t) = (f*g)(t) = \int_0^t f(\tau)g(t-\tau)d\tau$$ $$F(s) = \frac{1}{s^2} \implies f(t) = t$$ $$G(s) = \frac{1}{s^2-1} \implies g(t) = \sinh(t)$$ $$h(t) = t * \sinh(t) = \int_0^t\tau \sinh(t-\tau)d\tau$$

Answer 1

Hint.

Another way

$$ \frac{1}{s^2(s^2-1)}=\frac{A}{s^2}+\frac{B}{s-1}+\frac{C}{s+1} = \frac{(B+C)s^3+(A+B-C)s^2-A}{s^2(s^2-1)} $$ so we need

$$ \cases{B+C = 0\\ A+B-C=0\\ A=-1} $$

Regarding the convolution, as $f\circledast g = g\circledast f$ we have

$$ \int_0^t\tau \sinh(t-\tau)d\tau = \int_0^t(t-\tau) \sinh(\tau)d\tau $$

and

$$ \int_0^t(t-\tau) \sinh(\tau)d\tau = t\int_0^t\sinh(\tau)d\tau - \int_0^t\tau \sinh(\tau)d\tau $$

NOTE

$$ \sinh(x) = \frac 12(e^x-e^{-x}) $$