$$Let \space I=\int\limits_{x=0}^{x=1}\int\limits_{y=0}^{y=\sqrt{1+x^2}}\dfrac 1{\sqrt{(1+x^2)^2+y^2}}dxdy$$ $ \text{ Integrating w.r.t. y we get } \\ I=\int\limits_{x=0}^{x=1}\Bigg[\dfrac 1{\sqrt{1+x^2}}\tan^{-1}\dfrac y{\sqrt{1+x^2}}\Bigg]_0^{\sqrt{1+x^2}}dx\\ =\int\limits_0^1\dfrac 1{\sqrt{1+x^2}}\Bigg[\tan^{-1}\Bigg(\dfrac {\sqrt{1+x^2}}{\sqrt{1+x^2}}\Bigg)-\tan^{-1}(0)\Bigg]dx\\ =\int\limits_0^1\dfrac 1{\sqrt{1+x^2}}.\dfrac \pi4dx\\ =\dfrac \pi4[\log|x+\sqrt{1+x^2}|]^1_0\\ =\dfrac \pi4[\log|1+\sqrt2|-\log|0+\sqrt1|]\\ =\dfrac \pi4\log(1+\sqrt2) \because \log(1)=0$