= Solution
For $t>0$, integrate $e^{itz}/(1+z^2)$ over the upper semicircle of radius $R$. The arc tends to zero, since $|e^{itz}|\le1$ there and the denominator is of order $R^2$. The sole <pole> is $i$, of <residue> $e^{-t}/(2i)$, so the real part gives $I(t)=\pi e^{-t}$. The integrand defining $I$ is even in $t$, and $I(0)=\pi$ by the arctangent integral. Hence
$$
\boxed{I(t)=\pi e^{-|t|}\quad(t\in\mathbb R).}
$$
For $J$ use $e^{itz}/[z(1+z^2)]$ with a small clockwise semicircular indentation above zero. Its <residue> at zero is one, so the small arc tends to $-i\pi$. The <pole> $i$ has <residue> $-e^{-t}/2$, and the large arc again vanishes. Thus
$$
\operatorname{PV}\int_{-\infty}^\infty\frac{e^{itx}}{x(1+x^2)}dx-i\pi=-i\pi e^{-t}.
$$
The imaginary part is the ordinary convergent integral $J(t)$: its apparent singularity at zero is removable, with limit $t$, and its tails are absolutely integrable. Therefore $J(t)=\pi(1-e^{-t})$ for $t>0$. Oddness in $t$ and the value zero at $t=0$ give
$$
\boxed{J(t)=\begin{cases}\pi(1-e^{-t}),&t>0,\\0,&t=0,\\-\pi(1-e^{t}),&t<0.\end{cases}}
$$
<Differentiation> is valid at every real $t$, including zero. Indeed the <mean value theorem> bounds the absolute difference quotient of $\sin(tx)/[x(1+x^2)]$ with respect to $t$ by $1/(1+x^2)$, an integrable majorant independent of $t$. <Dominated convergence> therefore gives $\boxed{J'(t)=I(t)\text{ for all }t\in\mathbb R}$. In particular both one-sided derivatives at zero equal $\pi$.
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