Solve: $(y\sqrt{1-y^2})dx+(x\sqrt{1-y^2}+y)dy=0$ 
$$(y\sqrt{1-y^2})dx+(x\sqrt{1-y^2}+y)dy=0$$

$$\frac{\partial M}{\partial y}=\sqrt{1-y^2}-\frac{-2y^2}{2\sqrt{1-y^2}}=\frac{1-2y^2}{\sqrt{1-y^2}}$$
$$\frac{\partial N}{\partial x}= \sqrt{1-y^2}$$
$$\frac{M_{y}-N_{x}}{M}=\frac{\frac{1-2y^2}{\sqrt{1-y^2}}-\sqrt{1-y^2}}{y\sqrt{1-y^2}}=\frac{-y}{1-y^2}=h(y)$$
$$I=\exp{\left(-\int h(y) dy\right)}= \exp{\left(-\int\frac{-y}{1-y^2}dy\right)}=\exp{\left(\frac{-\ln|1-y^2|}{2}\right)}$$
Is there a way to simplify $$\exp{\left(\frac{-\ln|1-y^2|}{2}\right)}\;?$$

Apart from your question you can solve this equation by this way as well
$$(y\sqrt { 1-y^{ 2 } } )dx+(x\sqrt { 1-y^{ 2 } } +y)dy=0\\ ydx+xdy+\frac { ydy }{ \sqrt { 1-{ y }^{ 2 } }  } =0\\ d\left( xy \right) -\frac { 1 }{ 2 } d\left( \sqrt { 1-{ y }^{ 2 } }  \right) =0\\ d\left( xy-\frac { \sqrt { 1-{ y }^{ 2 } }  }{ 2 }  \right) =0\\ xy-\frac { \sqrt { 1-{ y }^{ 2 } }  }{ 2 } =C\\ \\ $$
