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.
 A: 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 $0.011$
A: Too long for a comment: It is known that $\displaystyle\int_0^\frac\pi2\sin(\sin x)dx=\int_0^\frac\pi2\sin(\cos x)dx=\frac\pi2H_0(1)$, 
and $\displaystyle\int_0^\frac\pi2\cos(\cos x)dx=\int_0^\frac\pi2\cos(\sin x)dx=\frac\pi2J_0(1)$, where H and J are the Struve and Bessel 
functions respectively, so it is not out of the question that the existence of such closed forms, at 
least in terms of these special functions, might have been one of the main initial triggers which 
determined our anonymous source to investigate the nature of $\displaystyle\int_0^\frac\pi2\!\tan(\tan x)dx$, which turns 
out to be $\ldots$ rather difficult, to say the very least. Indeed, the entire interval $\bigg(0,\dfrac\pi2\bigg)$ can be 
broken up into an infinite number of sub-intervals, the first one of which is $\bigg(0,~\arctan\dfrac\pi2\bigg)$, 
while all others are of the form $\bigg[\arctan\bigg(\dfrac\pi2+k\pi\bigg),~\arctan\bigg(\dfrac\pi2+\big(k+1\big)\pi\bigg)\bigg]$, for $k\in$ N. 
The integral diverges towards $+\infty$ on the former, and towards $-\infty$ on the rest, but in such 
a manner that $\displaystyle\int_0^{\alpha(k)}\!\tan(\tan x)dx\to\infty$ rather than $-\infty$, where $\alpha(k)=\arctan\bigg(\dfrac\pi2+k\pi\bigg)$. 
This can be better visualized by plotting the function's graphic. But determining whether the 
limit $\displaystyle\lim_{k\to\infty}\int_0^{\alpha(k)}\!\tan(\tan x)dx$ ultimately converges or diverges is by no means a trivial task 
$\big($indeed, it is quite daunting$\big)$, so, discouraged, either by the integral's probable divergence, or 
perhaps by its difficulty, or maybe by a bit of both, our anonymous source $\big($let's call him/her 
Person X$\big)$ decides to focus its attention now on $\bigg(0,~\arctan\dfrac\pi2\bigg)$, but not before changing the 
function into $~\dfrac1{\tan(\tan x)}$ , so as to avoid its divergence in $x=\arctan\dfrac\pi2$ . However, he now 
faces another problem: its divergence in $0$. In order to fix this new impediment, the function 
is further changed to $~\dfrac1{1+\tan(\tan x)}$ . Then, by a simple substitution, $\displaystyle\int_0^{\alpha(0)}\frac{dx}{1+\tan(\tan x)}$ 
becomes $\displaystyle\int_0^\frac\pi2\frac{dx}{(1+x^2)(1+\tan x)}$ , as already pointed out by ncmathsadist in the comment 
section above. That such an improvised integral would indeed have a meaningful closed form 
is no more likelier than for $\displaystyle\int_0^\frac\pi2\frac{dx}{1+\sin(\sin x)}=\int_0^\frac\pi2\frac{dx}{1+\sin(\cos x)}$ or $\displaystyle\int_0^\frac\pi2\frac{dx}{1+\cos(\cos x)}$ 
$=\displaystyle\int_0^\frac\pi2\frac{dx}{1+\cos(\sin x)}$ to possess one as well.
