# Help with finding $\int_0^\infty \frac{\arctan(x)\log(x^2+1)}{x(x^2+1)}$ [duplicate]

I am trying to evaluate $$\int_0^\infty \frac{\arctan(x)\log(x^2+1)}{x(x^2+1)}$$ I have tried differentiating under the integral sign, with different parameters, as well as contour integration, but I have not succeeded with neither of them. Wolfram Alpha gives a numerical value $$\approx 0.754694$$, but I suspect that there is a more exact answer.
## marked as duplicate by Namaste integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 19 '18 at 20:46
• This is solved here. See the middle of the accepted answer there "For the integral let $F(a) =\ldots$" – Lee David Chung Lin Oct 19 '18 at 19:04
• Exact value: $\frac{1}{2} \ln ^2(2) \pi$ – Mariusz Iwaniuk Oct 19 '18 at 19:28