= Solution
Use the branch from part (i) and close the real-axis path with a large upper <semicircle>, indenting clockwise around $-1$ and $1$. On the segment between them, take the upper-bank value of the <branch cut>. This is an instance of <square-root endpoint integrals by an upper-half-plane contour>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2-2016-ib-branch-contour.png]
{title=Upper-half-plane contour, clockwise endpoint indentations, and the two enclosed poles; indentation sizes are exaggerated}
The contributions can be controlled separately. Since $g(z)=O(z^{-2})$, the large arc contributes $O(R^{-1})\to0$. Near either <branch point>, $g=O(\varepsilon^{-1/2})$, so an arc of length $O(\varepsilon)$ contributes $O(\sqrt\varepsilon)\to0$. On the upper bank of $(-1,1)$,
$$
g(x+i0)=\frac{x(x^2+3)}{i(x^2+1)(x^2+4)\sqrt{1-x^2}},
$$
which is odd in $x$, so its integral is zero. On the exterior real intervals $h(-x)=-h(x)$, while the rational numerator is odd. Hence $g(-x)=g(x)$ and the two exterior contributions sum to $2J$.
The enclosed <residues> use $h(i)=i\sqrt2$ and $h(2i)=i\sqrt5$:
$$
\operatorname{Res}_{z=i}g=\frac{i(i^2+3)}{(2i)(i^2+4)h(i)}=-\frac{i}{3\sqrt2},\qquad
\operatorname{Res}_{z=2i}g=\frac{2i((2i)^2+3)}{((2i)^2+1)(4i)h(2i)}=-\frac{i}{6\sqrt5}.
$$
The <residue theorem> gives $2J=2\pi i[-i/(3\sqrt2)-i/(6\sqrt5)]$. Therefore
$$
\boxed{J=\frac\pi{3\sqrt2}+\frac\pi{6\sqrt5}.}
$$
As an independent check, $u=\sqrt{x^2-1}$ converts the integral to
$$
\int_0^\infty\frac{u^2+4}{(u^2+2)(u^2+5)}\,du=\int_0^\infty\left[\frac{2/3}{u^2+2}+\frac{1/3}{u^2+5}\right]du,
$$
which yields the same value. This check also confirms the branch signs and the factor of two in the contour argument.
Back to article page
