= Solution
Differentiate $x/(1+x^2)^n$ and integrate between zero and infinity. Its boundary values vanish for every integer $n\ge1$, giving
$$
0=\int_0^\infty\left[\frac1{(1+x^2)^n}-\frac{2nx^2}{(1+x^2)^{n+1}}\right]dx.
$$
Since $x^2/(1+x^2)^{n+1}=(1+x^2)^{-n}-(1+x^2)^{-n-1}$, this becomes $0=E_n-2n(E_n-E_{n+1})$. Hence the <recurrence relation> is
$$
\boxed{E_{n+1}=\frac{2n-1}{2n}E_n.}
$$
The starting value is $E_1=[\arctan x]_0^\infty=\pi/2$. Therefore $E_2=\pi/4$, $E_3=3\pi/16$, and
$$
\boxed{\int_0^\infty\frac{dx}{(1+x^2)^4}=E_4=\frac56\frac34\frac12\frac\pi2=\frac{5\pi}{32}.}
$$
These are also <Wallis integrals> after the <integration by substitution> $x=\tan t$.
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