A strange integral: ∫+∞−∞dx1+(x+tanx)2=π.\int_{-\infty}^{+\infty} {dx \over 1 + \left(x + \tan x\right)^2} = \pi.

Here is an approach.

We may use the following result, which goes back to G. Boole (1857) :

$$\int_{-\infty}^{+\infty}f\left(x-\frac{a_1}{x-\lambda_1}-\cdots-\frac{a_n}{x-\lambda_n}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \tag1$$

with $a_i>0, \lambda_i \in \mathbb{R}$ and $f$ sufficiently ‘regular’.

Observe that, for $x\neq n\pi$, $n=0,\pm1,\pm2,\ldots$, we have $$\cot x = \lim_{N\to +\infty} \left(\frac1x+\frac1{x+\pi}+\frac1{x-\pi}+\cdots+\frac1{x+N\pi}+\frac1{x-N\pi}\right)$$ leading to (see Theorem 10.3 p. 14 here and see achille’s answer giving a route to prove it)

$$\int_{-\infty}^{+\infty}f\left(x-\cot x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \tag2$$

with $\displaystyle f(x)=\frac{1}{1+\left(\small{\dfrac\pi2 -x }\right)^2}$.

On the one hand, from $(2)$,
$$\begin{align} \int_{-\infty}^{+\infty}f\left(x-\cot x\right)\mathrm{d}x& =\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \\\\ &=\int_{-\infty}^{+\infty}\frac{1}{1+\left(\small{\dfrac\pi2 -x }\right)^2}\: \mathrm{d}x\\\\ &=\int_{-\infty}^{+\infty}\frac{1}{1+x^2}\: \mathrm{d}x\\\\ & =\pi \tag3 \end{align}$$
On the other hand, with the change of variable $x \to \dfrac\pi2 -x$,
$$\begin{align} \int_{-\infty}^{+\infty}\!\!\!f\left(x-\cot x\right)\mathrm{d}x & =\int_{-\infty}^{+\infty} \!\!\!f\left(\dfrac\pi2-x-\tan x\right)\mathrm{d}x \\\\ & =\int_{-\infty}^{+\infty}\frac{1}{1+\left(x+ \tan x \right)^2} \mathrm{d}x \tag4 \end{align}$$
Combining $(3)$ and $(4)$ gives

$$\int_{-\infty}^{+\infty}\frac{1}{1+\left(x+ \tan x \right)^2} \mathrm{d}x=\pi.$$