TITLE: Finding $\int_{0}^{\frac{\pi}{4}} \frac{\tan \theta - \tan^3 \theta}{\ln \tan \theta}$
QUESTION [9 upvotes]: 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.

REPLY [7 votes]: $$\int_{0}^{1} \frac{1-x}{(1+x) \ln x}\,dx=-\int_0^1\int_0^1 \sum_{k=0}^{\infty}(-1)^k x^{y+k}\ dy \ dx=- \sum_{k=0}^{\infty}\int_0^1\int_0^1  (-1)^k x^{y+k}\ dx \ dy$$
$$- \sum_{k=0}^{\infty}(-1)^k \log\left(\frac{k+2}{k+1}\right)=-\log\left(\frac{2}{1}\cdot\frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdots\right)=-\log\left(\frac{\pi}{2}\right).$$