This can be done with a minimum of computation, relying only on (a) simple algebra and (b) basic knowledge of distributions associated with statistical tests. As such, the demonstration may have substantial pedagogical value--which is a fancy way of saying it's worth studying.
Let $Z=X/(X+Y),$ so that
$$Z - \frac{1}{2} = \frac{X}{X+Y} - \frac{X/2+Y/2}{X+Y} = \frac{1}{2}\frac{X-Y}{X+Y} = \frac{1}{2}\frac{(X-Y)/\sqrt{2}}{(X+Y)/\sqrt{2}} = \frac{1}{2}\frac{U}{V}$$
where $$(U,V) = \left(\frac{X-Y}{\sqrt{2}}, \frac{X+Y}{\sqrt{2}}\right).$$ Because $(U,V)$ is a linear transformation of the bivariate Normal variable $(X,Y),$ it too is bivariate Normal, and an easy calculation (ultimately requiring, apart from arithmetical definitions, only the fact that $1+1=2$) shows the variances of $U$ and $V$ are unity and $U$ and $V$ are uncorrelated: that is, $(U,V)$ also has a standard Normal distribution.
In particular, $U$ and $V$ are both symmetrically distributed (about $0$), implying $U/V$ has the same distribution as $U/|V|.$ But $|V| = \sqrt{V^2}$ has, by definition, a $\chi(1)$ distribution. Since $U$ and $V$ are independent, so are $U$ and $|V|,$ whence (also by definition) $U/|V| = U/\sqrt{V^2/1}$ has a Student t distribution with one degree of freedom.
The conclusion, after no integration and only the simplest of algebraic calculations, is
$W = 2Z-1 = U/V$ has a Student t distribution with one degree of freedom.
That's just another name for the (standard) Cauchy distribution. Since $Z = W/2 + 1/2$
is just a rescaled and shifted version of $W,$, $Z$ has a Cauchy distribution (once again by definition), QED.
Summary of facts used
Every one of the facts used in the foregoing analysis is of interest and well worth knowing.
These are basic theorems:
These are all definitions:
The odd integral moments have principal values of zero while the even integral moments have infinite principal values.
By definition, the Cauchy Principal Value [w.r.t. infinity] of an integral over the real line is
$$\operatorname{pv}\int_\mathbb{R} g(x)\,\mathrm{d}x = \lim_{R\to\infty} \int_{-R}^R g(x)\,\mathrm{dx}.$$
When $g$ is antisymmetric -- that is, $g(x) = -g(x)$ for all $x,$ obviously the right hand side is always zero, whence its limit is zero.
The Cauchy density is proportional to $1/(1+x^2),$ which is symmetric about $0.$ Consequently, for any antisymmetric integrable function $h,$
$$g(x) = \frac{h(x)}{1+x^2}$$
is also integrable and antisymmetric, whence
$$\operatorname{pv} \int_\mathbb R \frac{h(x)}{1+x^2}\,\mathrm{d}x = 0.$$
This is the case for $h(x) = x^{2k+1}$ when $k$ is a natural number. It is not the case when $h(x) = x^{2k}.$ With $k=0,$ the integral converges, but for $k\ge 1$ use the simple fact that when $|x|\ge 1,$
$$\frac{x^{2k}}{1+x^2} \ge \frac{x^{2k}}{x^2+x^2} = \frac{1}{2}x^{2k-2} \ge \frac{1}{2}$$ to estimate
$$\begin{aligned}
\lim_{R\to\infty}\int_{-R}^R \frac{x^{2k}}{1+x^2}\,\mathrm{d}x &\ge \lim_{R\to\infty}\left(\int_{-R}^{-1} + \int_1^R\right) \frac{x^{2k}}{1+x^2}\,\mathrm{d}x\\
&\ge \lim_{R\to\infty}\left(\int_{-R}^{-1} + \int_1^R\right) \frac{1}{2}\,\mathrm{d}x\\
&= \lim_{R\to\infty} R - 1 =\infty.
\end{aligned}$$
A similar argument by symmetry handles the case of $negative$ integral $n,$ but now the principal value is computed by excising a small neighborhood $(-\epsilon,\epsilon)$ around zero and taking the limit (from above) as $\epsilon\to 0.$ Again there's cancellation for odd $n$ but divergence for even $n.$ Alternatively, change variables from $x$ to $1/x,$ which reduces the integral to the form explicitly considered here.
Best Answer
As shown at How does entropy depend on location and scale?, the integral is easily reduced (via an appropriate change of variable) to the case $\gamma=1$, for which
$$H = \int_{-\infty}^{\infty} \frac{\log(1+x^2)}{1+x^2}\,dx.$$
Letting $x=\tan(\theta)$ implies $dx = \sec^2(\theta)d\theta$ whence, since $1+\tan^2(\theta) = 1/\cos^2(\theta)$,
$$H = -2\int_{-\pi/2}^{\pi/2} \log(\cos(\theta))d\theta = -4\int_{0}^{\pi/2} \log(\cos(\theta))d\theta .$$
There is an elementary way to compute this integral. Write $I= \int_{0}^{\pi/2} \log(\cos(\theta))d\theta$. Because $\cos$ on this interval $[0, \pi/2]$ is just the reflection of $\sin$, it is also the case that $I= \int_{0}^{\pi/2} \log(\sin(\theta))d\theta.$ Add the integrands:
$$\log\cos(\theta) + \log\sin(\theta) = \log(\cos(\theta)\sin(\theta)) = \log(\sin(2\theta)/2) = \log\sin(2\theta) - \log(2).$$
Therefore
$$2I = \int_0^{\pi/2} \left(\log\sin(2\theta) - \log(2)\right)d\theta =-\frac{\pi}{2} \log(2) + \int_0^{\pi/2} \log\sin(2\theta) d\theta.$$
Changing variables to $t=2\theta$ in the integral shows that
$$\int_0^{\pi/2} \log\sin(2\theta) d\theta = \frac{1}{2}\int_0^{\pi} \log\sin(t) dt = \frac{1}{2}\left(\int_0^{\pi/2} + \int_{\pi/2}^\pi\right)\log\sin(t)dt \\= \frac{1}{2}(I+I) = I$$
because $\sin$ on the interval $[\pi/2,\pi]$ merely retraces the values it attained on the interval $[0,\pi/2]$. Consequently $2I = -\frac{\pi}{2} \log(2) + I,$ giving the solution $I = -\frac{\pi}{2} \log(2)$. We conclude that
$$H = -4I = 2\pi\log(2).$$
An alternative approach factors $1+x^2 = (1 + ix)(1-ix)$ to re-express the integrand as
$$\frac{\log(1+x^2)}{1+x^2} = \frac{1}{2}\left(\frac{i}{x-i} + \frac{i}{x+i}\right)\log(1+ix) + \frac{1}{2}\left(\frac{i}{x-i} + \frac{i}{x+i}\right)\log(1-ix)$$
The integral of the first term on the right can be expressed as the limiting value as $R\to\infty$ of a contour integral from $-R$ to $+R$ followed by tracing the lower semi-circle of radius $R$ back to $-R.$ For $R\gt 1$ the interior of the region bounded by this path clearly has a single pole only at $x=-i$ where the residue is
$$\operatorname{Res}_{x=-i}\left(\left(\frac{i}{x-i} + \frac{i}{x+i}\right)\log(1+ix)\right) = i\left.\log(1 + ix)\right|_{x=-i} = i\log(2),$$
whence (because this is a negatively oriented path) the Residue Theorem says
$$\oint \left(\frac{1}{1+ix} + \frac{1}{1-ix}\right)\log(1+ix) \mathrm{d}x = -2\pi i (i\log(2)) = 2\pi\log(2).$$
Because the integrand on the circle is $o(\log(R)/R)$ which grows vanishingly small as $R\to\infty,$ in the limit we obtain
$$\int_{-\infty}^\infty \frac{1}{2}\left(\frac{1}{1+ix} + \frac{1}{1-ix}\right)\log(1+ix) \mathrm{d}x = \pi\log(2).$$
The second term of the integrand is equal to the first (use the substitution $x\to -x$), whence $H=2(\pi\log(2)) = 2\pi\log(2),$ just as before.