When I encountered the integral $\int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{2 } x\right)} d x $, I tried the substitution $x\mapsto \frac{1}{x} $ and found a wonderful result. $$I=\int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{2 } x\right)} d x \stackrel{x\mapsto\frac{1}{x}}{=} \int_0^{\infty} \frac{1}{x(1+x)\left(\pi+\ln ^{2 } x\right)} d x $$ Averaging them yields
$$ \begin{aligned} I & =\frac{1}{2} \int_0^{\infty}\left[\frac{1}{(1+x)\left(\pi+\ln ^2 x\right)}+\frac{1}{x(1+x)\left(\pi+\ln ^2 x\right)}\right]dx\\ & =\frac{1}{2} \int_0^{\infty} \frac{1}{x\left(\pi+\ln ^2 x\right)} d x \\ & =\frac{1}{2} \int_0^{\infty} \frac{1}{\pi+\ln ^2 x} d(\ln x) \\ & =\frac{1}{2 \sqrt{\pi}}\left[\tan ^{-1}\left(\frac{\ln x}{\sqrt{\pi}}\right)\right]_0^{\infty}\\ & =\frac{\sqrt{\pi}}{2} \end{aligned} $$
Then I want to generalize the result to the integral
$$I_n=\int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{2n } x\right)} d x $$ By replacing the power $2$ by $2n$, we have $$ \begin{aligned} I_n & =\frac{1}{2} \int_0^{\infty} \frac{1}{x\left(\pi+\ln ^{2 n} x\right)} d x\\ &= \frac{\sqrt[2 n]{\pi}}{2\pi} \int_{-\infty}^{\infty} \frac{1}{1+x^{2 n}} d x \quad (\textrm{ via } \ln x \mapsto \sqrt[2 n]{\pi} x) \end{aligned} $$ Using the well-known result $\int_0^{\infty} \frac{d x}{1+x^n}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)$, we have $$ I_n=\frac{\sqrt[2 n]{\pi}}{2 n} \csc \left(\frac{\pi}{2 n}\right) $$ For simplicity, given any even integer $m$, we have $$ \boxed{\int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{m } x\right)} d x =\frac{\sqrt[m]{\pi}}{m} \csc \left(\frac{\pi}{m}\right)} $$
For examples, $$ \begin{aligned} & \int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{2 } x\right)} d x =\frac{\sqrt{\pi}}{2} \\ & \int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{4 } x\right)} d x =\frac{\sqrt[4]{\pi}}{2 \sqrt{2}} \\ & \int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{6 } x\right)} d x =\frac{\sqrt[6]{\pi}}{3} \end{aligned} $$
Are there any other methods? Comments and alternative methods are highly appreciated.
Here is a more general extension
$$I_n=\int_0^{\infty} \frac{1}{x(1+x^a)\left(\pi+\ln ^{2n } x\right)}\ dx =\frac{\pi^{\frac1{2n}}}{2n\sin\frac\pi{2n}}$$