6.23 Example continued

We determine $F(x)$ using partial fractions. We have

so $F(x)={\log|x-1|-\log|x+1|=}\,{\log\left|\frac{x-1}{x+1}\right|.}$ Hence $e^{F(x)}={\left|\frac{x-1}{x+1}\right|.}$ In this process we can essentially ignore the absolute value sign (which at worst means multiplying by $(-1)$), so we can take for the integrating factor: $I(x)=\frac{x-1}{x+1}$.

Multiplying both sides through by $I(x)$, we obtain the equation

where the left-hand side is ${\frac{d}{dx}\left(\frac{x-1}{x+1}y\right).}$