10. Evaluate
$$
\int_{-\infty}^{\infty} \frac{1-x^{2}}{1+x^{4}} d x
$$
Solution: $\square$
Let $S=\int_{0}^{\infty} 1 /\left(x^{4}+1\right) d x$; note that the integral converges absolutely. Substituting $x=1 / u$, so that $d x=-1 / u^{2} d u$, we have
$$
\begin{array}{c}
S=\int_{0}^{\infty} \frac{1}{1+x^{4}} d x=\int_{\infty}^{0} \frac{1}{1+u^{-4}} \frac{d u}{-u^{2}}=\int_{\infty}^{0} \frac{-u^{2}}{u^{4}+1} d u \\
=\int_{0}^{\infty} \frac{u^{2}}{1+u^{4}} d u=\int_{0}^{\infty} \frac{x^{2}}{1+x^{4}} d x
\end{array}
$$
(the manipulations are justified by absolute convergence), from which we see that $\int_{0}^{\infty}\left(1-x^{2}\right) /\left(1+x^{4}\right) d x=0$. Since the integrand is an even function, it follows that the integral from $-\infty$ to $\infty$ is zero as well.