<s>[INST] 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 $$\exp{\left(\frac{-\ln|1-y^2|}{2}\right)}\;?$$
 [/INST] $a\ln b=\ln b^a\implies -\frac{1}{2}\ln \left |1-y^2\right |=$ $\ln \frac{1}{\sqrt{1-y^2}}$ and $e^{\ln u}=u$  integrating factor is going to be $\frac{1}{\sqrt{1-y^2}}$
</s>