A novelty integral for $\pi$

The integrand can be broken up as $$I=\int_0^{\infty} \left(\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x} +\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln x}-\frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x}\right)dx.$$

But, by integration by parts, $$\int \frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x} dx= \int \frac{2\ln(1+x)}{x^{1/2} \ln x} d\left(\ln\left(\frac{1+x}{2}\right)\right) \\=\small\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{1/2} \ln x}-2\int \left( \frac{\ln\left(\frac{1+x}{2}\right)}{x^{1/2} (1+x) \ln x}-\frac{ \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x}-\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{2 x^{3/2} \ln x}\right)dx$$

That is, $$\int \left(\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x} +\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln x}-\frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x}\right)dx \\= -\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{1/2} \ln x}+2 \int \frac{\ln\left(\frac{1+x}{2}\right)}{x^{1/2} (1+x) \ln x} dx$$

And the claim follows from Jack D'Aurizio's preliminary result.

A preliminary result.

$$I_1=\int_{0}^{+\infty}\frac{2\log\left(\frac{1+x}{2}\right)}{\sqrt{x}(1+x)\log(x)}=\color{red}{\pi}.\tag{1}$$

Proof: through the substitution $x=e^t$, $$\begin{eqnarray*}I_1=\int_{-\infty}^{+\infty}\frac{\log\left(\frac{e^t+1}{2}\right)}{\cosh\left(\frac{t}{2}\right)t}\,dt&=&\color{red}{\frac{1}{2}\int_{-\infty}^{+\infty}\frac{dt}{\cosh\left(\frac{t}{2}\right)}}+\color{blue}{\int_{-\infty}^{+\infty}\frac{\log\cosh\left(\frac{t}{2}\right)}{t\cosh\left(\frac{t}{2}\right)}\,dt}\end{eqnarray*}$$ where the red integral is easy to compute and the blue one vanishes by symmetry.

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Thanks to nospoon, this is ultimately everything we need to prove the OP's identity through integration by parts.