$$\int_0^{\infty} \arctan{\left(\frac{n}{\cosh{(x)}}\right)} \mathop{dx}$$ I think the integral evaluates to $$\frac{\pi}{2} \ln{\left(\sqrt{n^2+1}+n\right)}$$ but I dont know how really! I think $n$ is any number but I dont know for sure! The answer reminds me of $\int \frac{\pi}{2} \sec{x} \mathop{dx}$ and $n=\tan{x}$.
I got to $$\int_0^{\infty} \arctan{\left(\frac{e^{x} n}{e^{2x}+1}\right)} \mathop{dx}$$ $$\int_0^{\infty} \arctan{\left(\frac{n}{2}\frac{e^{x} +e^x}{e^{x}\cdot e^x+1}\right)} \mathop{dx}$$ Reminds me of $\tan{a-b}$ but the $n/2$ factor?
from Hot Weekly Questions - Mathematics Stack Exchange
Saelee
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