With the branch cut on $[0,+\infty)$, on the lower edge of the slit the integrand becomes $$\frac{\log x + 2\pi i}{(i^3x^{3/4})(1+x)}$$
since the value of $z^{3/4}$ changes by a factor of $\exp \frac{3\cdot 2\pi i}{4} = i^3$ for each winding around the origin. Thus, taking the orientation into account,
$$\frac{-2\pi^2}{\exp \frac{3\pi i}{4}} = (1 - i)\int_0^{+\infty} \frac{\log x}{x^{3/4}(1+x)}\,dx + 2\pi \int_0^{\infty} \frac{dx}{x^{3/4}(1+x)}\,.$$
Taking the imaginary part on both sides gives the result.
We can use some instances of Schroder's Integral, which evaluates Gregory coefficients, namely:
$$(-1)^{n-1}G_n=\int_0^\infty\frac{1}{(\pi^2+\ln^2 t)(1+t)^n}dt;\quad G_1=\frac12, \ G_2=-\frac{1}{12}$$
But let's adjust things first.
$$I=\int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} dx \overset{x^2=t}=2\int_0^\infty \frac{(\pi\sqrt t-\ln t)^3}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$=2\pi^3\int_0^\infty \frac{t}{(\pi^2+\ln^2 t)(1+t)^2}dt-6\pi^2 \int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$+6\pi \int_0^\infty \frac{\ln^2 t}{(\pi^2+\ln^2 t)(1+t)^2}dt-2\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$=2\pi^3 I_1-6\pi^2I_2+6\pi I_3-2 I_4$$
$$I_1=\int_0^\infty \frac{t}{(\pi^2+\ln^2 t)(1+t)^2}dt=\int_0^\infty \frac{(1+t)-1}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$=\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)}dt-\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$=G_1+G_2=\frac12-\frac{1}{12}=\frac{5}{12}$$
The $I_2$ integral can be found here.
$$I_2=\int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt\overset{t=\frac{1}{x}}=-\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=\frac{\pi}{24}$$
$$I_3=\int_0^\infty \frac{\ln^2 t}{(\pi^2+\ln^2 t)(1+t)^2}dt = \int_0^\infty \frac{(\pi^2 +\ln^2 t)-\pi^2}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$=\int_0^\infty \frac{1}{(1+t)^2}dt-\pi^2\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$=1+\pi^2 G_2=1-\frac{\pi^2}{12}$$
$$I_4=\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}=\int_0^\infty \frac{\ln t((\pi^2+\ln^2 t ) -\pi^2 )}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$=\int_0^\infty \frac{\ln t}{(1+t)^2}\frac{dt}{\sqrt t}-\pi^2 \int_0^\infty \frac{\ln t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$=-\pi+\pi^2 I_2=-\pi +\frac{\pi^3}{24}$$
$$\require{cancel} \Rightarrow I=\cancel{2\pi^3{\frac{5}{12}}} -\cancel{6\pi^2{\frac{\pi}{24}}}+6\pi \left({1-\cancel{\frac{\pi^2}{12}}}\right)-2\left({-\pi +\cancel{\frac{\pi^3}{24}}}\right)={8\pi}$$
Best Answer
HINT:
Note that we have for $a>\varepsilon$, $|\varepsilon^2e^{i2\phi}+a^2|\ge a^2 -\varepsilon^2$ and
$$ \lim_{\varepsilon\to0} \frac{2\pi \varepsilon \log^2(\varepsilon)}{a^2-\varepsilon^2} =0 $$