# Evaluation of given indefinite integral

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?

• Does it really have a closed form solution? Feb 6 '17 at 16:46

We have \begin{aligned} \left[\frac{2-x^2}{(1-x)\sqrt {1-x^2}} \right] &= \left[\frac {1-x^2}{(1-x)\sqrt {1-x^2}} + \frac {1}{(1-x)\sqrt {1-x^2}}\right] \\ &= \left[\sqrt {\frac {1+x}{1-x}} + \frac{1}{(1-x)\sqrt {1-x^2}}\right] \\ &= \left[f (x)+g (x)\right] \end{aligned}

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

\begin{aligned} \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}} \\ \end{aligned} The only thing left is to notice that,I'll leave that to you. $$\left(\frac{\sqrt{1+x}}{\sqrt{1-x}}\right)'=\frac{1}{(1-x)\sqrt{1-x^2}}$$

HINT:

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

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

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

$$\mathrm d\left(\dfrac{u}v\right) =\dfrac{u'v-uv'}{v^2}$$

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