$$\begin {aligned} \int _{0}^{1}\frac {\arcsin \sqrt x}{1+x}dx &=\int _{0}^{1}\frac {2x\arcsin x}{1+x^{2}}dx\\ &=\left [\ln (1+x^{2})\arcsin x\right ]_0^1-\int _{0}^{1}\frac {\ln (1+x^{2})}{\sqrt {1-x^{2}}}dx\\ &=\frac {\pi \ln 2}2-\int _{0}^{\frac {\pi }2}\ln \left (1+\sin ^{2}\theta \right )d\theta \\ &=\frac {\pi \ln 2}2-\int _{0}^{\frac {\pi }2}\left (\left [\ln (t^{2}+\sin ^{2}\theta )\right ]_{t=0}^1+\ln \sin ^{2}\theta \right )d\theta \\ &=\frac {\pi \ln 2}2-\int _{0}^{\frac {\pi }2}\int _{0}^{1}\frac {2t}{t^{2}+\sin ^{2}\theta }dtd\theta -2\int _{0}^{\frac {\pi }2}\ln \sin \theta d\theta \\ &=\frac {\pi \ln 2}2-\int _{0}^{1}\int _{0}^{\frac {\pi }2}\frac {2t}{t^{2}+\sin ^{2}\theta }d\theta dt+\pi \ln 2\\ &=\frac {3\pi \ln 2}2-\int _{0}^{1}\int _{0}^{\frac {\pi }2}\frac {2t}{t^{2}(1+\tan ^2\theta )+\tan ^{2}\theta }\frac {d\theta}{\cos ^{2}\theta } dt\\ &=\frac {3\pi \ln 2}2-\int _{0}^{1}\int _{0}^{\infty }\frac {2t}{t^{2}+(1+t^2)s^{2}}dsdt\\ &=\frac {3\pi \ln 2}2-\int _{0}^{1}\frac {\pi }{\sqrt {1+t^{2}}}dt\\ &=\textcolor {blue}{\frac {3\pi \ln 2}2-\pi \ln (1+\sqrt 2)}. \end {aligned}$$