{\displaystyle
\int\frac{\hbox{d}x}{(1+x^2)\sqrt{1-x^2}}=\\
\int\frac{\frac{4t}{(t^2+1)^2}}{\frac{2(t^4+1)}{(t^2+1)^2}\cdot \frac{-2t}{t^2+1}}\hbox{d}t=\\
-\int\frac{t^2+1}{t^4+1}\hbox{d}t=\\
-{1\over 2}\int\Bigg (\frac{1}{t^2+1+\sqrt{2}t}+\frac{1}{t^2+1-\sqrt{2}t}\Bigg)\hbox{d}t=\\
-{1\over \sqrt{2}}\Bigg(\newcommand{\arctg}{\mathop{\mathrm{arctg}}}{\arctg(\sqrt{2}t+1)+\arctg(\sqrt{2}t-1)}\Bigg)+D
}