Solve $\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$ 
Find $$\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$$

Looking at the numerator, combined with the surd, you can get $$\int\frac{e^x(1+x)\left(\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}\right)}{1-x^2}\mathrm dx$$Then this begins to look like the quotient rule, since the denominator is $\sqrt{1-x^2}^2$, and in the numerator, $1/\sqrt{1-x^2}$ is almost like the derivative of the square root. However, it isn't quite that - there are extra terms. $$\color{red}{e^x\left(\sqrt{1-x^2}+\frac{x}{\sqrt{1-x^2}}\right)}+\frac{e^x}{\sqrt{1-x^2}}+e^x x\sqrt{1-x^2}$$
The red terms are accounted for by quotient rule (giving an integral of $\frac{e^x}{\sqrt{1-x^2}}$). But then what do we do with the remaining terms? 

Hint
Observe that the exponent of $1-x$ is $-\dfrac32$
So, let us find $$\dfrac{d\left(e^x\dfrac{(1+x)^n}{\sqrt{1-x}}\right)}{dx}$$
Compare with the given expression to find the value of $n$
