Hi I'm trying to show that $$ I=\int_0^1 \frac{\arctan\big(\sqrt{x^2 + 2}\big)}{\sqrt{x^2 + 2}(x^2 + 1)}dx=\frac{5\pi^2}{96}. $$ We can try the substitution $u=(x^2+2)^{1/2}, du=x(2+x^2)^{1/2}dx$ but that didn't help me much because of the (x^2+1) piece. Any ideas? Thanks.
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@RandomVariable Nice. Thank you again – Jeff Faraci Apr 16 '14 at 17:16

See also the book "Inside Interesting Integrals..." – Yuriy S Mar 21 '16 at 08:12