= Solution
{c}
For $a\ne b$, integrate
$$
F(z)=\frac{e^{iz}}{(z^2+a^2)(z^2+b^2)}
$$
along the <contour> consisting of $[-R,R]$ and the upper <semicircle>, with <positive orientation of a contour> and $R>\max\{a,b\}$. Its enclosed <simple poles> are $ia$ and $ib$, with <residues>
$$
\operatorname{Res}(F,ia)=\frac{e^{-a}}{2ia(b^2-a^2)},\qquad
\operatorname{Res}(F,ib)=\frac{e^{-b}}{2ib(a^2-b^2)}.
$$
On the upper <semicircle>, $|e^{iz}|=e^{-\operatorname{Im}z}\leq1$. The <reverse triangle inequality> gives $|z^2+a^2|\geq R^2-a^2$ and $|z^2+b^2|\geq R^2-b^2$, so the <estimation lemma> gives
$$
\left|\int_{\mathrm{arc}}F(z)\,dz\right|\leq\frac{\pi R}{(R^2-a^2)(R^2-b^2)}\longrightarrow0.
$$
The <integral> along the <real axis> is absolutely convergent: the denominator stays nonzero on bounded real <intervals>, and the integrand is $O(x^{-4})$ at infinity. The <residue theorem> therefore gives
$$
\int_{-\infty}^{\infty}\frac{e^{ix}}{(x^2+a^2)(x^2+b^2)}\,dx
=\frac{\pi}{a^2-b^2}\left(\frac{e^{-b}}b-\frac{e^{-a}}a\right).
$$
Take <real parts>. The real integrand is an <even function>, so
$$
\boxed{\int_0^\infty\frac{\cos x}{(x^2+a^2)(x^2+b^2)}\,dx
=\frac{\pi}{2(a^2-b^2)}\left(\frac{e^{-b}}b-\frac{e^{-a}}a\right).}
$$
To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\leq b\leq3a/2$, the <absolute value> of the integrand is bounded by
$$
\frac1{(x^2+a^2)(x^2+a^2/4)},
$$
an integrable <function> on $[0,\infty)$. The <dominated convergence theorem> justifies passing the limit under the <integral>. If $h(t)=e^{-t}/t$, then $h'(a)=-e^{-a}(a+1)/a^2$, and
$$
\lim_{b\to a}\frac{h(b)-h(a)}{a^2-b^2}=-\frac{h'(a)}{2a}=\frac{e^{-a}(a+1)}{2a^3}.
$$
Hence
$$
\boxed{\int_0^\infty\frac{\cos x}{(x^2+a^2)^2}\,dx=\frac{\pi e^{-a}(a+1)}{4a^3}.}
$$
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