Evaluate integral $\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx $ I tried to solve it by the following method:
\begin{align*}
    \int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx&= \int \frac{(1-x^2)e^x+\frac{e^x}{2}+\frac{e^x}{2}}{(1-x)(\sqrt{1-x^2}) }dx \\ &= \int \frac{(1-x^2)e^x+\frac{(1-x)e^x}{2}+\frac{(1+x)e^x}{2}}{(1-x)(\sqrt{1-x^2}) }dx \\ &=\int \frac{\sqrt{(1-x^2)}e^x+\frac{\sqrt{(1-x)}e^x}{2\sqrt{x+1}}+\frac{\sqrt{(1+x)}e^x}{2\sqrt{1-x}}}{(1-x) }dx \\&=
    \int \frac{\sqrt{1-x}(\sqrt{x+1}\cdot e^x+\frac{e^x}{2\sqrt{x+1}})+\sqrt{x+1}\cdot e^x \cdot \frac{1}{2\sqrt{1-x}}}{1-x}dx \\&= \int\left(\frac{\sqrt{x+1}\cdot e^x}{\sqrt{1-x}}\right)'dx \\ &=\left(\frac{\sqrt{x+1}\cdot e^x}{\sqrt{1-x}}\right)+C
\end{align*}
Is this solution right?
if it is right are there other methods to solve it.

Recognize
$$\frac{2-x^2}{(1-x)(\sqrt{1-x^2})}= \frac{(1-x^2)+1}{(1-x)(\sqrt{1-x^2})}\\=\sqrt{\frac{1+x}{1-x} }+ \frac{1}{(1-x)^{3/2}\sqrt{1+x})}
= \sqrt{\frac{1+x}{1-x} }+ \left( \sqrt{\frac{1+x}{1-x} }\right)’
$$
and then apply to $f(x)= \sqrt{\frac{1+x}{1-x} }$ the general integral result
$$\int e^x(f(x)+f’(x)) dx= e^x f(x)$$
to obtain
$$\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx 
=\sqrt{\frac{1+x}{1-x} }\>e^x +C
$$
