Finding the value of $\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta$. It is given that $$
I=\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta=\pi\ln k-\frac{\pi^2}{w}
$$ and was asked to find the value $kw$. Here is my try on it:
Substituting $\theta = \dfrac{\pi}{4}+x$, we get
$$\begin{equation} \label{eq1}
\begin{split}
I &= \displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{(-4x) \tan (\frac{\pi}{4}+x)}{1-\tan (\frac{\pi}{4}+x)} \, dx \\
  &= -4\displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{x.\dfrac{1+ \tan x}{1- \tan x}}{1- \dfrac{1+ \tan x}{1- \tan x}} \, dx \\
&= 2 \displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{x(1+\tan x)}{\tan x} \, dx
\end{split}
\end{equation}$$
From here I am stuck on this, I don't know how to proceed from here to get answer in terms of $\pi \ln k-\frac{\pi^2}{w}$
Thank you for any help.

Note that
\begin{align}
\int_0^{\pi/2}\frac x{\tan x}dx 
=&\int_0^{\pi/2}\int_0^1 \frac1{1+y^2\tan^2 x}dy \ dx
=\frac\pi2\int_0^1\frac1{1+y}dy=\frac\pi2\ln2
\end{align}
and
\begin{split}
I= &\ 2 \int_{-\frac{\pi}{2}}^0\dfrac{x(1+\tan x)}{\tan x} \, dx
= 2 \int^{\frac{\pi}{2}}_0\left(\frac x{\tan x}- x\right) dx
= \pi \ln2 -\frac{\pi^2}4
\end{split}
