Integral $I=\int_0^1 \frac{\arctan\big(\sqrt{x^2 + 2}\big)}{\sqrt{x^2 + 2}(x^2 + 1)}dx$

Question

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.