Unsure about solution to $\int_0^\frac{\pi}{4} \frac{\ln\left| 1 + \tan(x)\right|}{\left( 1 + \tan(x)\right)^n}\:dx$ Spurred on by a question posed on MSE, I was hoping to resolve the following definite integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{4} \frac{\ln\left| 1 + \tan(x)\right|}{\left( 1 + \tan(x)\right)^n}\:dx
\end{equation}
Where $n \in \mathbb{N},\: n\geq 2$. The approach I've taken is valid (I believe), but the final solution seems invalid. Here I used employ Feynman's Trick by introducing the following function:
\begin{equation}
 J(p) = \int_0^\frac{\pi}{4} \frac{\ln\left|1 + \tan(x)\right|}{p + \tan(x)}\:dx
\end{equation} 
Where $p \in \mathbb{R}$, $ 0\leq p \leq 1$. We observe that:
\begin{equation}
I_n = \frac{(-1)^{n - 1} J^{n - 1}(1)}{(n - 1)!}
\end{equation}
Where $J^m(p)$ is the $m$-th derivative of $J(p)$. To resolve $J(p)$ we first let $u = \tan(x)$ to yield:
\begin{align}
J(p) &= \int_0^1 \frac{\ln\left|1 + u\right|}{\left(u + p\right)\left(u^2 + 1\right)}\:du = \int_0^1 \frac{\ln\left|1 + u\right|}{p^2 + 1}\left[ \frac{1}{u + p} + \frac{p}{u^2 + 1} - \frac{u}{u^2 + 1} \right]\:du \nonumber \\
&=\frac{1}{p^2 + 1}\left[A(p) + pB - C \right]
\end{align}
Now $B$ and $C$ can be resolved without non-elementary functions using the self-similar substitution:
\begin{equation}
B = \int_0^1 \frac{\ln\left| u + 1\right|}{u^2 + 1}\:du = \frac{\pi}{8}\ln(2), \qquad C = \int_0^1 \frac{u}{u^2 + 1}\ln\left| u + 1\right|\:du = \frac{\pi^2}{96} + \frac{\ln^2(2)}{8}
\end{equation}
$A(p)$ can be resolved into the Dilogarithm with some simple substitions and repositioning. We first let $w = u + p$ to yield:
\begin{align}
A(p) &= \int_0^1 \frac{\ln\left|1 + u\right|}{u + p}\:du = \int_p^{p + 1} \frac{\ln\left| 1 + w - p \right|}{w}\:dw = \int_p^{p + 1} \frac{\ln\left|\left(1 - p\right)\left( \frac{w}{1 - p}  + 1\right) \right|}{w}\:dw \nonumber \\
&= \ln\left|1 - p\right|\int_p^{p + 1} \frac{1}{w}\:dw + \int_p^{p + 1} \frac{\ln\left|\frac{w}{1 - p}  + 1\right|}{w}\:dw \nonumber  \\
&= \ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right| + \int_p^{p + 1} \frac{\ln\left|\frac{w}{1 - p}  + 1\right|}{w}\:dw 
\end{align} 
For the final integral, let $u = \frac{w}{1 - p}$:
\begin{align}
A(p) &= \ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right| + \int_{\frac{p}{1 - p}}^{\frac{1 + p}{1 - p}} \frac{\ln\left|u  + 1\right|}{u}\:du \nonumber \\
&= \ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right| + \bigg[ -\operatorname{Li}_{2}(-u)\bigg]_{\frac{p}{1 - p}}^{\frac{ 1 + p}{1 - p}} \nonumber \\
&= \ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right|  + \left[ \operatorname{Li}_{2}\left(\frac{p}{p - 1}\right) - \operatorname{Li}_{2}\left(\frac{p + 1}{p - 1}\right) \right]
\end{align} 
Thus $J(p)$ becomes:
\begin{align}
J(p) &= \frac{1}{p^2 + 1}\bigg[\ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right|  + \left[ \operatorname{Li}_{2}\left(\frac{p}{p - 1}\right) - \operatorname{Li}_{2}\left(\frac{p + 1}{p - 1}\right) \right]  \nonumber \\
&\quad + \frac{\pi}{8}\ln(2)p- \left(\frac{\pi^2}{96} + \frac{\ln^2(2)}{8}\right) \bigg]
\end{align}
My concern is evaluating this at $p = 1$. Have I fallen prey to an invalid use of the Linearity property of continuous functions? Is my method valid?
 A: This is not an answer but a different method to find a closed form of ${{I}_{n}}$ 
For $x={{\tan }^{-1}}\left( u \right)$  we have:
$$
{{I}_{n}}=\int_{0}^{1}{\frac{\ln \left( 1+u \right)}{{{\left( 1+u \right)}^{n}}\left( 1+{{u}^{2}} \right)}du}
$$
Now using this result (thanks to Sangchul Lee) :
$$
\frac{1}{{{(1+x)}^{n}}(1+{{x}^{2}})}=\left( \sum\limits_{k=1}^{n}{\frac{\sin (k\pi /4)}{{{2}^{k/2}}}}\frac{1}{{{(x+1)}^{n+1-k}}} \right)+\frac{\cos (n\pi /4)-x\sin (n\pi /4)}{{{2}^{n/2}}(1+{{x}^{2}})}
$$
Multiply both sides by $\ln \left( 1+x \right)$ and integrate from $0$ to $1$ you get (separate the last term in the sum):
$$
\begin{align}
  & {{I}_{n}}=\frac{\sin (n\pi /4)}{{{2}^{n/2}}}\int_{0}^{1}{\frac{\ln \left( 1+x \right)dx}{(x+1)}}+\sum\limits_{k=1}^{n-1}{\left[ \frac{\sin (k\pi /4)}{{{2}^{k/2}}}\int_{0}^{1}{\frac{\ln \left( 1+x \right)dx}{{{(x+1)}^{n+1-k}}}} \right]} \\ 
 & \quad +\frac{\cos (n\pi /4)}{{{2}^{n/2}}}\int_{0}^{1}{\frac{\ln \left( 1+x \right)dx}{(1+{{x}^{2}})}}-\frac{\sin (n\pi /4)}{{{2}^{n/2}}}\int_{0}^{1}{\frac{x\ln \left( 1+x \right)dx}{(1+{{x}^{2}})}} \\ 
 &  \\ 
\end{align}
$$
at this point we have every thing except:
$$
\int_{0}^{1}{\frac{\ln \left( 1+x \right)dx}{{{(x+1)}^{n+1-k}}}}=\frac{1+{{2}^{k-n}}\left( k-n \right)\ln 2-{{2}^{k-n}}}{{{\left( k-n \right)}^{2}}},\quad k<n$$
Finally
$$
\begin{align}
  & {{I}_{n}}={{\ln }^{2}}2\frac{\sin (n\pi /4)}{{{2}^{n/2+1}}}+\sum\nolimits_{k=0}^{n-1}{\left[ \frac{\sin (k\pi /4)}{{{2}^{k/2}}}\frac{1+{{2}^{k-n}}\left( k-n \right)\ln 2-{{2}^{k-n}}}{{{\left( k-n \right)}^{2}}} \right]} \\ 
 & \quad +\pi \ln 2\frac{\cos (n\pi /4)}{{{2}^{n/2+3}}}-\frac{\sin (n\pi /4)}{{{2}^{n/2}}}\left( \frac{{{\pi }^{2}}}{96}+\frac{{{\ln }^{2}}2}{8} \right) \\ 
\end{align}
$$
