Evaluate $ \int_{0}^{\frac{\pi}2}\frac1{(1+x^2)(1+\tan x)}\:\mathrm dx$ Evaluate the following integral
$$
\tag1\int_{0}^{\frac{\pi}{2}}\frac1{(1+x^2)(1+\tan x)}\,\mathrm dx
$$



My Attempt:

Letting $x=\frac{\pi}{2}-x$ and using the property that

$$
\int_{0}^{a}f(x)\,\mathrm dx = \int_{0}^{a}f(a-x)\,\mathrm dx
$$

we obtain

$$
\tag2\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{\left(1+\left(\frac{\pi}{2}-x\right)^2\right)(1+\tan x)}\,\mathrm dx
$$

Now, add equation $(1)$ and $(2)$. After that I do not understand how I can proceed further.
Here is an approach.  

We give some preliminary results.



   The poly-Hurwitz zeta function



The poly-Hurwitz zeta function may  initially be defined  by the series
$$
\begin{align}
  \displaystyle \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, 
  \quad \Re a &gt;-1, \, \Re b &gt;-1, \, \Re s&gt;0. \tag1
\end{align} 
$$
This special function is a natural extension of the Hurwitz zeta function initially defined as
$$
\zeta(s,a)=\sum_{n=0}^{\infty} \frac{1}{(n+a)^s}, \quad \Re a&gt;0, \Re s&gt;1, \tag2
$$ 
which is a natural extension itself of the Riemann zeta function initially defined as
$$
\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}, \quad \Re s&gt;1. \tag3
$$

The poly-Hurwitz function appears in different places with different notations, one may find it here:
 [Masri, p. 2 and p. 15  (2004)], [Murty, p. 17 (2006)], [Sinha, p. 45 (2002)]. In this answer we are dealing with a simplified version of a general poly-Hurwitz function. 

The series in $(1)$ converges absolutely for $\displaystyle \Re s&gt;0$. Moreover, the convergence of the series is uniform on every
half-plane $$\displaystyle H_{\delta}=\left\{s \in \mathbb{C}, \Re s \geq  \delta \right\}, \, \delta \in \mathbb{R},\, \delta&gt;0,$$ therefore the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot \mid a,b)$
is analytic on the half-plane $\displaystyle \Re s&gt;0$.

Let $a$, $b$ and $s$ be complex numbers such that $\Re a &gt;-1, \, \Re b &gt;-1, \, \Re s &gt;0$. One may observe that
$$
\begin{align}
  \zeta(s\mid a,b) &amp; =  \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}\\
 &amp; = \sum_{n=1}^{+\infty} \frac{(n+b)+(a-b)}{(n+a)^{s+1}(n+b)}\\
 &amp; = \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}}+(a-b)\sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}(n+b)} \tag4
\end{align} 
$$
giving the functional identity
$$
\begin{align}
 \zeta(s \mid a,b) = \zeta(s+1,a+1) +(a-b) \zeta(s+1 \mid a,b) \tag5
\end{align} 
$$

where $\displaystyle \zeta(\cdot,\cdot)$ is the standard Hurwitz zeta function.

From $(5)$, we obtain by induction, for $n=1,2,3,\ldots $,

$$
\begin{align}
 \zeta(s\mid a,b) = \sum_{k=1}^{n}(a-b)^{k-1}\zeta(s+k,a+1) +(a-b)^n\zeta(s+n \mid a,b). \tag6
\end{align} 
$$

We use $(6)$ to extend $\displaystyle \zeta(\cdot \mid a,b)$ to a meromorphic function on each open set $\Re s&gt;-n $, $n\geq 1$. Since the Hurwitz zeta function is analytic on the whole complex plane except for a simple pole at $1$ with residue $1$, then from $(6)$ the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot\mid a,b)$ is analytic on the whole complex plane except for a simple pole at $0$ with residue $1$.



   The poly-Stieltjes constants



In 1885 Stieltjes has found  that the Laurent series expansion around $1$ of the Riemann zeta function
$$
\zeta(1+s) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k s^k, \quad s \neq 0,\tag7
$$  is such that the coefficients of the regular part of the expansion are given by
$$
\begin{align} 
\gamma_k&amp; = \lim_{N\to \infty}\left(\sum_{n=1}^N \frac{\log^k n}{n}-\frac{\log^{k+1} \!N}{k+1}\right).
\end{align} \tag8
$$ Euler was the first to define a constant of this form (1734)
$$
\begin{align} 
\gamma &amp; = \lim_{N\to\infty}\left(1+\frac12+\frac13+\cdots+\frac1N-\log N\right)=0.577215\ldots.
\end{align} 
$$ The constants $\displaystyle \gamma_k$ are called the Stieltjes constants and due to the fact that $\displaystyle \gamma_0=\gamma$   they are sometimes called the generalized Euler's constants.

Similarly, Wilton (1927) and Berndt (1972) established that the  Laurent series expansion in the  neighbourhood of $1$ of the Hurwitz zeta function 
$$
\begin{align} \zeta(1+s,a) 
= \frac1s+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a)\:s^{k}, \quad \Re a&gt;0, \,s\neq 0, \tag{9}
\end{align}
$$ is such that the coefficients of the regular part of the expansion are given by
$$
\begin{align} 
\gamma_k(a)&amp; = \lim_{N\to \infty}\left(\sum_{n=0}^N \frac{\log^k (n+a)}{n+a}-\frac{\log^{k+1} (N+a)}{k+1}\right), \quad \Re a&gt;0,
\end{align} \tag{10}
$$ 
with $\displaystyle \gamma_{0}(a)=-\psi(a)=-\Gamma'(a)/\Gamma(a)$. The coefficients $\gamma_k(a)$ are called the generalized Stieltjes constants.

We have seen from $(6)$ that the poly-Hurwitz zeta function admits a Laurent series expansion around $0$. Let's denote by $\displaystyle\gamma_k(a,b)$ the coefficients of the regular part of $\displaystyle \zeta(\cdot\mid a,b)$ around $0$. I will call these coefficients the poly-Stieltjes constants. 

Do we have an analog of $(10)$ for $\displaystyle\gamma_k(a,b)$?

The following result is new.


  Theorem 1. Let $a,b$ be complex numbers such that $\Re a &gt;-1, \, \Re b &gt;-1$. Consider 
  The poly-Hurwitz zeta function $$
\begin{align}
   \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, 
  \quad \Re s&gt;0. \tag{11}
\end{align} 
$$
  Then the meromorphic extension of $\displaystyle \zeta(\cdot\mid a,b)$ admits the following Laurent series expansion around $0$,
  $$
\zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{+\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k, \quad s \neq 0,\tag{12}
$$ where the poly-Stieltjes constants $\displaystyle \gamma_k(a,b)$ are given by $$
\begin{align} 
\gamma_k(a,b)&amp; = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right)
\end{align} \tag{13}
$$ with $$ \gamma_{0}(a,b)=-\psi(b+1)=-\Gamma'(b+1)/\Gamma(b+1). \tag{14}$$


Proof. Let $a,b$ be complex numbers such that $\Re a &gt;-1, \, \Re b &gt;-1$.

We first assume $\Re s&gt;0$. Observing that, for each $n \geq 1$,
$$
\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right| \leq \sum_{k=0}^{\infty}\left|\frac{\log^k(n+a)}{n+b}\right|\frac{|s|^k }{k!}&lt;\infty
$$ and that
$$
\sum_{n=1}^{\infty}\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right|=\sum_{n=1}^{\infty}\left|\frac1{(n+a)^s(n+b)}\right| = 
\sum_{n=1}^{\infty}\frac1{|n+a|^{\Re s}|n+b|}&lt;\infty,$$
we obtain
$$
\begin{align}
&amp;\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right)  s^k \\\\
&amp;= \lim_{N\to+\infty}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) s^k \\\\
&amp;=\lim_{N\to+\infty}\sum_{k=0}^{\infty}\left(\sum_{n=1}^N\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\
&amp;=\lim_{N\to+\infty}\left(\sum_{n=1}^N\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\
&amp;=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{(n+a)^s(n+b)} +\frac1{N^s}-\frac1s\right) \\\\
&amp;=\zeta(s \mid a,b)-\frac1{s}
\end{align}
$$  as desired. Then, using $(6)$, we extend the preceding identity by analytic continuation  to all $s \neq 0$. To prove $(14)$, we start from a standard series representation of the digamma function (see Abram. &amp; Steg. p. 258 6.3.16):
$$
\begin{align}
-\psi(b+1) &amp;= \gamma - \sum_{n=1}^{\infty} \left( \frac1n - \frac1{b+n}  \right) \\
&amp;=\lim_{N\to+\infty}\left(\gamma - \sum_{n=1}^N\left( \frac1n - \frac1{b+n}  \right)\right)\\
&amp;=\lim_{N\to+\infty}\left(\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)-\left(\sum_{n=1}^N\frac1n-\ln N-\gamma \right)\right)\\
&amp;=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)\\\\
&amp;=\gamma_0(a,b)
\end{align} 
$$ using $(13)$.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$



One of the consequences of Theorem 1 is the new possibility to express some series in terms of the poly-Stieltjes constants.


  Theorem 2. Let $a,b,c$ be complex numbers such that $\Re a &gt;-1, \, \Re b &gt;-1, \, \Re c &gt;-1$. 
  
  Then 
  $$
\begin{align}
  (b-a)\sum_{n=1}^{+\infty} \frac{\log (n+c)}{(n+a)(n+b)}=\gamma_1(c,a)-\gamma_1(c,b), \tag{15}
\end{align} 
$$ similarly
  $$
\begin{align}
 \sum_{n=1}^{+\infty} \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)=\gamma_1(a,b)-\gamma_1(c,b), \tag{16}
\end{align} 
$$ with the poly-Stieltjes constant
  $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). 
$$ 


Proof. Let $a,b,c$ be complex numbers such that $\Re a &gt;-1, \, \Re b &gt;-1, \, \Re c &gt;-1$.

We have
$$
 (b-a)\frac{\log (n+c)}{(n+a)(n+b)}=\frac{\log (n+c)}{n+a}-\frac{\log (n+c)}{n+b}
$$ giving, for $N\geq1$,
$$
\begin{align}
  (b-a)&amp;\sum_{n=1}^N \frac{\log (n+c)}{(n+a)(n+b)}=\\\\
&amp;  \left(\sum_{n=1}^N\frac{\log (n+c)}{n+a}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{17}
\end{align}
$$
letting $N \to \infty$  and  using $(13)$ gives $(15)$.

We have, for $N\geq1$,
$$
\begin{align}
  &amp;\sum_{n=1}^N \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)\\\\
&amp;=  \left(\sum_{n=1}^N\frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{18}
\end{align}
$$
letting $N \to \infty$ and  using $(13)$ gives $(16)$.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$



   Juantheron's integral



Let’s first give a numerical evaluation of Juantheron’s integral.

I would like to thank Jonathan Borwein and David H. Bailey who obtained the result below to $1000$ digits, in just 3.9 seconds run time, using David's new MPFUN-MPFR software, along with the tanh-sinh quadrature program included with the MPFUN-MPFR  package. 

They also tried the integral with David's MPFUN-Fort package, which has a completely different underlying multiprecision system, and they obtained the same result below. 

Finally, they computed the integral with Mathematica $11.0$;  it agreed with the result below, although it required about 10 times longer to run. 


  Proposition 1. We have
  $$
\begin{align}  \int_{0}^{\Large\frac{\pi}2}\!&amp;\frac1{(1+x^2)(1+\tan x)}\mathrm dx\\\\=
0.&amp;59738180945180348461311323509087376430643859042555\\
&amp;67307703207161550311033249824121789098990404474443\\
&amp;73300942847961727020952797366230453350097928752529\\
&amp;62099371263365268445580755896768905606293308536674\\
&amp;89639352215352393870280616186538538722285601087082\\
&amp;81730013060929540132583577240799018025603130403772\\
&amp;83596189879605956759516344861849456740112012597646\\
&amp;30195536341071109827787231788650530475635336662512\\
&amp;50757672078320586388500276160658476344052492489409\\
&amp;64026178233152015087197531148322444147655936720008 \tag{19}\\
&amp;40650450631581050321100329502169853063154902765446\\
&amp;58804861176982696627707544105655815406116180984371\\
&amp;54148587721902800400109013880620460529382772599713\\
&amp;06874977209651994186527207589425408866256042399213\\
&amp;80515694164361264997143539392018681691584285790381\\
&amp;65536517701019826846772718498479534803417547866296\\
&amp;23842162877309354675086691711521468623807334908897\\
&amp;71491673168051054009130049879837629516862688171756\\
&amp;13790927986073268994254629238035029442300668334396\\
&amp;901581838911515359223628586133156893962372426055\cdots
\end{align}
$$


David H. Bailey confirmed that Mathematica $11.0$, in spite of the great numerical precision, could not find a closed-form of the integral.

The next result proves that the OP integral admits a closed form in terms of the poly-Stieltjes constants.


  Proposition 2. We have
  $$
\begin{align}  \int_{0}^{\Large\frac{\pi}2}\!\!\frac1{(1+x^2)(1+\tan x)}\mathrm dx
&amp;=\frac{(e^2+1)^2}{2(e^4+1)}\arctan \! \frac{\pi}{2}-\frac{e^4-1}{4(e^4+1)}\log\left(1+\frac{\pi^2}{4}\right)\\\\
&amp;+\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\
&amp;+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{20}\\\\ &amp;+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right)
\end{align}
$$ with the poly-Stieltjes constant 
  $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$


Proof. We  proceed in three steps.

Step 1. One may write
$$
\require{cancel}
\begin{align}
&amp;\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\mathrm dx \\
&amp;=\int_{0}^{\Large\frac{\pi}{2}}\frac{\cos x}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\
&amp;=\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{(\cos x+\sin x)+(\cos x-\sin x)}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\
&amp;=\frac12\int_{0}^{\Large\frac{\pi}{2}}\!\frac{1}{1+x^2}\mathrm dx+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)}\frac{(\cos x-\sin x)}{(\cos x+\sin x)}\mathrm dx\\
&amp;=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{1+x^2}\tan (x-\pi/4)\:\mathrm dx\\
&amp;=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{-\Large\frac{\pi}{4}}^{\Large\frac{\pi}4}\frac{1}{1+(x+\pi/4)^2}\tan x \:\mathrm dx\\
&amp;=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_0^{\Large\frac{\pi}4}\left(\frac1{1+(x+\pi/4)^2}-\frac1{1+(x-\pi/4)^2}\right)\tan x \:\mathrm dx\\
&amp;=\frac12 \arctan\! \frac{\pi}{2}-\frac{\Im}2 \!\int_{0}^{\Large\frac{\pi}{4}}\!\!\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx \tag{21}
\end{align}
$$
Let’s evaluate the latter integral.

Step 2. One may recall that the tangent function, as a meromorphic function, can be expressed as an infinite sum of rational functions:
$$
\tan x = \sum_{n=0}^{+\infty} \frac{2x}{\pi^2 (n+1/2)^2-x^2}, \quad x \neq \pm \pi/2, \pm 3\pi/2,\pm 5\pi/2,\ldots. \tag{22}
$$
We have the inequality
$$
\sup_{x \in [0,\pi/4]}\left|\frac{2x}{\pi^2 (n+1/2)^2-x^2}\right|\leq \frac1{(n+1/2)^2}, \quad n=0,1,2,\ldots, \tag{23}
$$ the convergence in $(22)$ is then uniform on  $[0,\pi/4]$. Thus, plugging $(22)$ into $(21)$, we are allowed to integrate $(21)$ termwise.  

Each term, via a partial fraction decomposition, is evaluated to obtain
$$
\begin{align}
\int_{0}^{\Large\frac{\pi}4}\!&amp;\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right)\frac{2x}{\pi^2 (n+1/2)^2-x^2}\:\mathrm dx\\
&amp;=\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\
&amp;+\frac1{\pi}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\
&amp;+\frac1{\pi}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)
\end{align} \tag{24}
$$
where for the sake of convenience we have set $\tau:=\pi/4+i$.

Step 3. We sum $(24)$ from $n=0$ to $\infty$ obtaining

$$
\begin{align}
\int_{0}^{\Large\frac{\pi}{4}}\!&amp;\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\
&amp;=\sum_{n=0}^{\infty}\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\
&amp;+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\
&amp;+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right), \tag{25}
\end{align}
$$ then, singling out the first terms in the two last series and using Theorem $2$ $(16)$, we get
$$
\begin{align}
\int_{0}^{\Large\frac{\pi}{4}}\!&amp;\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\
&amp;=\tan \tau \log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)
+\frac{4\pi}{\pi^2 -4\tau^2}\log 3
\\&amp;+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 +\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 +\frac{\tau}{\pi}\!\right)\tag{26}\\ &amp;+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 -\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 -\frac{\tau}{\pi}\!\right)
\end{align} $$ and the substitution $\tau=\pi/4+i$ gives the desired result.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$



Achille Hui's conjecture is true. 

Achille Hui has announced in the comments that the OP integral is equal to 
$$
\begin{align}
\frac{\arctan(\frac{\pi}{2}) - t\log\sqrt{1+\frac{\pi^2}{4}}}{1+t^2}+\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) } \tag{27}
\end{align}
$$ with $\displaystyle t := \tanh(1)$.

The first term in $(27)$, with a little algebra is seen  to be equal to the sum of the first two terms on the right hand side of $(20)$.

We establish the veracity of the conjecture using Proposition $2$ and using the next result.


  Proposition 3. We have
  $$
\begin{align}  &amp;\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) }\\\\
&amp;=\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\
&amp;+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{28}\\\\ &amp;+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right)
\end{align}
$$ with the poly-Stieltjes constant 
  $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right).$$


Proof. Observe that the first term of the series on the left hand side of  $(28)$, given by $n=0$, is just equal to
$$
\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)}.
$$ By a partial fraction decomposition, one may check that
$$
\begin{align}
\frac{\pi^2}4 &amp;\frac{ (2n+1) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) }
=\frac{\Im}{2\pi}\left(\!\frac1{n+\frac34+\frac{i}{\pi}}-\frac1{n+\frac14+\frac{i}{\pi}}\!\right) \tag{29}
\end{align}
$$ 
then, multiplying $(29)$ by $\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right)\right)$ and summing from $n=1$ to $\infty$ we get, using Theorem $2$ $(16)$, the result $(28)$.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$
A good fit for $\frac{1}{1+\tan x}$ in $[0,\frac{\pi}{2}]$ is $1-\frac{2}{\pi}x$

So

$$I_{approx}=\int_0^{\frac{\pi}{2}}\frac{1-\frac{2}{\pi}x}{1+x^2}dx=\arctan \frac{\pi}{2}-\frac{1}{\pi}\ln\left (1+ \frac{\pi^2}{4} \right )$$
Absolute error is about $0.011$