The original question was to evaluate:
$$\int_{0}^{1} \frac{1-x}{(1+x) \ln x}\,dx$$
Using the substitution $x=\tan^2 \theta $, I simplified it down to the integral $\int_{0}^{\frac{\pi}{4}} \frac{\tan \theta - \tan^3 \theta}{\ln \tan \theta}$.
From here, I am stuck and am not sure where to continue.
Since: $$ \int_{0}^{1}\frac{x^n-x^{n+1}}{\log x}\,dx = \log\left(\frac{n+2}{n+1}\right) \tag{1}$$ by Frullani's theorem, by expanding $\frac{1-x}{1+x}$ as its Taylor series in zero, collecting terms of the form $(x^n-x^{n+1})$ and integrating termwise, we have that our integral just equals minus the logarithm of the Wallis product, hence $\color{red}{-\log(\pi/2)}$ as claimed.
$$\int_{0}^{1} \frac{1-x}{(1+x) \ln x}\,dx=-\int_0^1\int_0^1 \sum_{k=0}^{\infty}(-1)^k x^{y+k}\ dy \ dx=- \sum_{k=0}^{\infty}\int_0^1\int_0^1 (-1)^k x^{y+k}\ dx \ dy$$ $$- \sum_{k=0}^{\infty}(-1)^k \log\left(\frac{k+2}{k+1}\right)=-\log\left(\frac{2}{1}\cdot\frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdots\right)=-\log\left(\frac{\pi}{2}\right).$$
