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Evaluate the given integral

$$\int e^x \bigg[\frac{2-x^2}{(1-x)\sqrt{1-x^2}} \bigg]dx$$

I was trying to convert it to $\int e^x (f(x)+f'(x))dx=e^x \cdot f(x)+C$ but did not succeed in algebraic manipulations. Could someone hint me to something so that I could proceed?

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  • $\begingroup$ Does it really have a closed form solution? $\endgroup$ – Math-fun Feb 6 '17 at 16:46
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$$\frac{2-x^2}{(1-x)\sqrt{1-x^2}}=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{1-x^2}{(1-x)\sqrt{1-x^2}}=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{\sqrt{1-x^2}}{(1-x)}=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{\sqrt{1+x}}{\sqrt{1-x}}$$ The only thing left is to notice that,I'll leave that to you. $$(\frac{\sqrt{1+x}}{\sqrt{1-x}})'=\frac{1}{(1-x)\sqrt{1-x^2}}$$

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We have $$[\frac{2-x^2}{(1-x)\sqrt {1-x^2}} ] = [\frac {1-x^2}{(1-x)\sqrt {1-x^2}} + \frac {1}{(1-x)\sqrt {1-x^2}}] = [\sqrt {\frac {1+x}{1-x}} + \frac{1}{(1-x)\sqrt {1-x^2}}] = [f (x)+g (x)] $$

Now take the derivative of $f (x)$ and compare it with $g(x)$. Hope it helps.

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HINT:

As $(1-x)\sqrt{1-x^2}=\sqrt{1+x}(1+x)^{3/2},$

$\int\dfrac1{\sqrt{1+x}}dx=2\sqrt{1+x}$

$\int\dfrac1{\sqrt{1-x}^{3/2}}dx=-\dfrac2{\sqrt{1-x}}$

$\dfrac{d(u/v)}{dx}=\dfrac{u'v-uv'}{v^2}$

start with $\dfrac{d\sqrt{\dfrac{1+x}{1-x}}}{dx}$

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