Central binomial coefficient from mathworld
$$\frac{2^{2n+1}}{\pi}\int_{0}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx={2n\choose n}$$
Here we have $\ln{2n\choose n}$ in term of another integral,
$$\int_{0}^{\infty}\frac{e^{-2nx}+2nx-1}{x(e^x+1)}dx=\ln{2n\choose n}$$
Can anyone verify this integral, thank you.
One may observe that $$ \int_0^\infty \frac{1}{(e^x+1)}dx=\int_0^\infty \frac{e^{-x}}{(1+e^{-x})}dx=\int_0^1 \frac{du}{1+u}=\ln2, \tag1 $$ combining it with the following result $$ \int_0^\infty \frac{1-e^{-sx}}{x(e^x+1)}dx=\frac12\log(\pi)+\log \Gamma\left(1+\frac{s}2\right)-\log \Gamma\left(\frac{1+s}2\right), \quad s>0, \tag2 $$ proved here, setting $s:=2n$, one gets
$$ \int_{0}^{\infty}\frac{e^{-2nx}+2nx-1}{x(e^x+1)}dx=\log\left(\frac{2^{2n}\:\Gamma\left(n+\frac12\right)}{\sqrt{\pi} \:\Gamma(n+1)} \right), \quad n>0, \tag3 $$
then, as $n=0,1,2\ldots,$ one may conclude with the duplication formula: $$ \frac{2^{2n}\:\Gamma\left(n+\frac12\right)}{\sqrt{\pi} \:\Gamma(n+1)}=\frac{(2n)!}{(n!)^2}=\binom{2n}{n} $$ which gives the announced result.
