Laplace transform of $$\frac{2\sin(t) \sinh(t)}{t}$$
Using the property:
$$\int_0^\infty \frac{f(t)}{t}dt = \int_0^\infty L(f(t)) du$$
where laplace transform is written as function of $u$, we get:
$$\int_0^\infty \frac{1}{(u-1+s)^2+1}-\frac{1}{(u+1+s)^2+1}du\\
= \arctan(s+1)+\arctan(1-s)$$
which gives $\arctan(\frac{2}{s^2-1})$ but answer given is $\arctan(\frac{2}{s^2})$.
Best Answer
Your $\arctan(s+1)+\arctan(1-s)$ is correct, but that does not give $\arctan(2/(s^2-1))$:
$$\tan(\arctan(s+1)+\arctan(1-s)) = \frac{2}{1-(s-1)(-s-1)} = \frac{2}{s^2}$$