[math-fun] An improper integral equals to log(\pi/2)-\gamma

Can anyone tell me how to prove that \int_0^\infty log(x)/(1+cosh(x)) dx = log(\pi/2)-\gamma? Any info related to this integral is also appreciated. I could find nothing except that Wolfram|Alpha knew it. Hope that this isn't too trivial for math-fun. Thanks, Warut