Consider

> 
> $$\lim\_{n\to\infty}\ln{n}-\int\_{0}^{n}{e^x-x-1\over x(e^x+1)}=\ln{\pi}-\gamma\tag1$$
> 
> 
> 

How does one show that?

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[Wolfram integrator](http://www.wolframalpha.com/input/?i=integrate%20(e%5Ex-x-1)%2F(x(e%5Ex%2B1))%20dx) can't evaluate the indefinite integral $(1)$