Inverse Laplace Transform, need help with this $\frac{s^2}{s^2+\sqrt{2}s+1}$

Question

Inverse Laplace Transform of $\frac{s^2}{s^2+\sqrt{2}s+1}$

I transformed the denominator as $(s+\frac{\sqrt{2}}{2})^{2}$ + $\frac{1}{2}$

$\frac{s^2}{(s+\frac{\sqrt{2}}{2})^{2} + \frac{1}{2}}$ and I have no idea how to move forward because partial fraction decomposition fails.

Answer 1

Hint:

$$\frac{s^2}{s^2+\sqrt2\,s+1}=1-\frac{\sqrt2\left(s+\frac{\sqrt2}2\right)}{\left(s+\frac{\sqrt2}2\right)^2+\frac12}$$

Answer 2

Substitute $u=s+\dfrac{\sqrt{2}}{2}$: $$\dfrac{s^2}{(s+\dfrac{\sqrt{2}}{2})^{2} + \frac{1}{2}}=\dfrac{(u-\dfrac{\sqrt{2}}{2})^2}{u^{2} + \dfrac{1}{2}}$$ Then expand...And apply the first Shifting Theorem.

$$\mathcal{L^{-1}}\dfrac {\left(s+\frac{\sqrt2}2\right)}{{\left(s+\frac{\sqrt2}2\right)^2+\frac12}}=e^{-\frac {\sqrt 2}{2}t}\cos \left(\dfrac {\sqrt 2}{2}t \right)$$