Finding the inverse Laplace transform of $ \ln \! \left( 1 + \frac{1}{s^{2}} \right) $.

The idea is to undo the operations you may find in the transform using the properties of the Laplace transform. The Laplace transform of the logarithm times Heaviside is essentially a logarithm divided by $s$. The transform of a derivative is essentially the transform of the function times $s$. And finally the transform of a function multiplied by an exponential is the transform of the function translated.

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We can write

$$\begin{align}\ln\left(1+\frac{1}{s^2}\right)&=\ln(s+i)+\ln(s-i)-2\ln(s)\\&=(s+i)\frac{\ln(s+i)+\gamma}{s+i}+(s-i)\frac{\ln(s-i)+\gamma}{s-i}-2s\frac{\ln(s)+\gamma}{s}\end{align}$$

We know that $$\mathcal{L}(-\ln(t)u(t))=\frac{\ln(s)+\gamma}{s}$$

Therefore

$$\mathcal{L}\left(D\left(-\ln(t)u(t)\right)\right)=s\frac{\ln(s)+\gamma}{s}$$

where $D$ is derivative and

$$\mathcal{L}\left(e^{\pm it}D\left(-\ln(t)u(t)\right)\right)=(s\pm i)\frac{\ln(s\pm i)+\gamma}{s\pm i}$$

Now you can complete it.