# 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.

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• Your result is perfect. Only one comment from me : very good work and $\to +1$ Jan 5, 2021 at 13:54

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 $$\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx =\sqrt{\frac{1+x}{1-x} }\>e^x +C$$