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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?

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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 $n$

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  • $\begingroup$ Wow, that was surprisingly simple... With $n=1/2$, I suppose this means my method with the quotient rule was bound to fail! $\endgroup$ – John Doe Apr 4 at 1:55
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Hint:

$$\dfrac{1+1-x^2}{(1-x)^{3/2}(1+x)^{1/2}}=\dfrac1{...}+f(x)$$

where $f(x)=\dfrac{\sqrt{1+x}}{\sqrt{1-x}},$

$f'(x)=?$

Recall $\dfrac{d(e^xf(x))}{dx}=?$

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$x=\cos2t,dx=?$

$$-I=\int\dfrac{e^{\cos2t}(1+\sin^22t)}{\sin^2t}=e^{\cos2t}\csc^2t-\dfrac{d(e^{\cos2t})}{dt}(-\cot t)$$

$$=\dfrac{d(e^{\cos2t}(-\cot t))}{dt}$$

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