Hello am looking for a solution to proving this.
$$
I:=\int_0^\infty \log(1+x^2)\frac{\cosh \pi x +\pi x\sinh \pi x}{\cosh^2 \pi x}\frac{dx}{x^2}=4-\pi.
$$
This one is related to Integral $\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}$ that the community together seemed to solve!
I tried writing
$$
I=\int_0^\infty \log(1+x^2)\frac{dx}{x^2\cosh \pi x}+\int_0^\infty \log(1+x^2)\frac{\pi x\tanh \pi x}{\cosh \pi x }\frac{dx}{x^2}
$$
but is not so clear now that I have two integrals to solve. I wasn't sure how integrating by parts would give me a clearer integral as the terms do not clean up here like the last one. I am not sure how else
Introducing something like $I(\alpha), I'(\alpha)$ in this situation did help but not much after this: $$ I(\alpha)=\int_0^\infty \log(1+\alpha x^2) \frac{\cosh \pi x +\pi x\sinh \pi x}{\cosh^2 \pi x}\frac{dx}{x^2}, \frac{dI}{d\alpha}= $$ $$ \int_0^\infty \frac{dx}{1+\alpha x^2}\frac{\cosh \pi x +\pi x\sinh \pi x}{\cosh^2 \pi x}=\int_0^\infty \frac{dx}{(1+\alpha x^2)\cosh \pi x}+\pi\int_0^\infty \frac{dx}{1+\alpha x^2}\frac{x\tanh \pi x}{\cosh \pi x} $$ Thank you