Evaluate $\displaystyle \int_{0}^{1}{\frac{\arctan \left( \sqrt{2+x^{2}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}.$
I make use of (again) 'magic differentiation' to solve this problem .
Consider $\displaystyle \int_0^1 \frac{dx}{(x^2+1)\sqrt{x^2+2} } $
Sub. $\displaystyle x = \frac{1}{t} $ , it becomes
$\displaystyle \int_1^{\infty} \frac{t~dt}{(t^2+1)\sqrt{1+2t^2} }$
sub $\displaystyle 1+2t^2 = u^2 $ .... After a few steps , we can obtain
$\displaystyle \int_0^1 \frac{dx}{(x^2+1)\sqrt{x^2+2} } = \frac{\pi}{6}$
The integral
$\displaystyle \int_{0}^{1}{\frac{\arctan \left( \sqrt{2+x^{2}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$
$\displaystyle = \int_{0}^{1}{\frac{\frac{\pi}{2} - \arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$
$\displaystyle = \frac{\pi}{2} \int_0^1 \frac{dx}{(x^2+1)\sqrt{x^2+2} } - \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$
$\displaystyle = \frac{\pi^2}{12} - \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$
Consider $\displaystyle I(a) = \int_{0}^{1}{\frac{\arctan \left( \frac{a}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx} $
Magic Differentiation !
$\displaystyle I'(a) = \int_0^1 \frac{dx}{(x^2+1)(x^2 + a^2 + 2 )} $
$\displaystyle = \frac{1}{1 + a^2 } \int_0^1 \left[ \frac{1}{1 + x^2 } - \frac{1}{x^2 + a^2+2} \right]~dx $
$\displaystyle = \frac{\pi}{4}\cdot \frac{1}{1+a^2} - \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+a^{2}}} \right)}{\left( 1+a^{2} \right)\sqrt{2+a^{2}}}\,dx}$
So $\displaystyle \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx} = I(1) = I(1)-I(0)$
$\displaystyle = \frac{\pi^2}{16}- I(1) $
$\displaystyle = \frac{\pi^2}{32} $
The answer to the problem is $\displaystyle \frac{\pi^2}{12} - \frac{\pi^2}{32} $
$\displaystyle = \frac{5 \pi^2}{96} $