Integral $\int_{0}^{\infty} \frac{\left(\frac{\pi x}{2}-\log (x)\right)^3}{\left(x^2+1\right)^2 \left(\log^2(x)+\frac{\pi ^2}{4}\right)} = \pi$ proof

Let $$f(x)= \frac{\pi^3 x^3-6 \pi ^2 x^2 \log (x)-3 \pi ^3 x+2 \pi ^2 \log (x)}{2 \left(x^2+1\right)^2 \left(4\log ^2(x)+\pi ^2\right)} $$ that is to say $$f(x)=\frac{\pi ^2 \left(\pi x \left(x^2-3\right)-2\left(3x^2-1\right) \log (x)\right)}{2 \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}$$ and write $$I=\int_0^\infty f(x)\,dx=\int_0^1 f(x)\,dx+\int_1^\infty f(x)\,dx$$ For the second integral, let $x=\frac 1 x$. So, now we have $$I=\int_0^1 f(x) \,dx-\int_0^1 g(x) \,dx=\int_0^1 [f(x)- g(x)] \,dx$$ with $$g(x)=\frac{\pi ^2 \left(\pi \left(3 x^2-1\right)+2 x \left(x^2-3\right) \log (x)\right)}{2 x \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}$$ $$f(x)-g(x)=\frac{\pi ^2 \left(\pi \left(x^4-6 x^2+1\right)-8 x \left(x^2-1\right) \log (x)\right)}{2 x \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}$$ and, numerically, $$\int_0^1 [f(x)- g(x)] \,dx=0$$ which still needs to be proved.

In fact, what happens if that $$\int_0^a [f(x)- g(x)] \,dx=-\int_a^1 [f(x)- g(x)] \,dx$$ where $a$ is the solution of $$\pi \left(x^4-6 x^2+1\right)-8 x \left(x^2-1\right) \log (x)=0$$ $$a=0.1928617994587428536765548799193482174970163765555\cdots$$ which is not recognized by inverse symbolic calculators.