I need to compute the answer of $ \frac{d y}{d x}=\frac{1}{1-x^{2}} \cdot y+(1+x)$
I have tried Integrating factor method and I get
$y=e^{\int_{0}^{x} \frac{1}{1-x^{2}} d{x}}\left(1+\int_{0}^{x}(1+x) e^{-\int_{0}^{x} \frac{1}{1-x^{2}} d x} d x\right)$ $=\sqrt{\left|\frac{1+x}{1-x}\right|}\left(1+\int_{0}^{x}(x+1) \sqrt{\mid\frac{1-x}{1+x}\mid} d x\right)$
But I can’t remove the absolute value. So anyone can help me?