Integration Bee

$\begin{aligned} I & = \int_0^\infty \frac {\ln (1+x^n) - \ln 2}{(1+x^2)\ln x} dx & \small \color{#3D99F6} \text{Using }\int_0^\infty f(x) \ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2} dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln (1+x^n) - \ln 2}{(1+x^2)\ln x} + \frac {\ln \left(1+ \frac 1{x^n}\right) - \ln 2}{x^2\left(1+\frac 1{x^2} \right)\ln \frac 1x} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln (1+x^n) - \ln 2}{(1+x^2)\ln x} + \frac {\ln \left(\frac {x^n+1}{x^n}\right) - \ln 2}{x^2\left(\frac {x^2+1}{x^2} \right) (-\ln x)} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln (1+x^n) - \ln 2}{(1+x^2)\ln x} - \frac {\ln (1+x^n) - \ln x^n - \ln 2}{(1+x^2)\ln x} \right) dx \\ & = \frac 12 \int_0^\infty \frac {\ln x^n}{(1+x^2) \ln x} dx = \frac 12 \int_0^\infty \frac {n \ln x}{(1+x^2) \ln x} dx \\ & = \frac 12 \int_0^\infty \frac n{1+x^2} dx = \frac n2 \tan^{-1} x \ \bigg|_0^\infty = \frac {n\pi}4 \end{aligned}$

Therefore, $\dfrac {n\pi}4 = 2\pi \implies n = \boxed 8$ .

hint1: let $t = \frac{1}{x}$

hint2: $\int_{0}^{\infty} \frac{n \ln t - \ln(1+t^n) + \ln2}{(1+t^2)\ln t} dt$

hint3: $\frac{1}{4} n \pi = 2\pi$