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So, the question says I have to perform the indefinite integration
$$\int\frac{e^x(2-x^2)}{(1-x)\sqrt{1-x^2}}dx$$ I know that
$$\int e^x(f(x)+f'(x))dx=e^xf(x)+C$$ Since any other substitution (using $x=ln(t)$ etc.) doesn't work, I expect I have to break the fraction with $e^x$ in the above integral to make separate fractions for $f(x)$ and $f'(x)$ somehow. I separate $2-x^2=1+(1-x^2)$, but that doesn't work (leaves $x$ in the numerator of derivative which I don't see in the question). Any other tricks?

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    $\begingroup$ $f(x)$ is missing on the right-hand side. $\endgroup$ – joriki Feb 23 '13 at 7:54
  • $\begingroup$ I don't believe to much in Wolfram, but he said that there is no result found in terms of standards mathematical function $\endgroup$ – Cortizol Feb 23 '13 at 13:59
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Your trick does work.

$$e^x\frac{2-x^2}{(1-x)\sqrt{1-x^2}}=e^x\frac1{(1-x)\sqrt{1-x^2}}+e^x\sqrt{\frac{1+x}{1-x}}$$

And, $$\frac{d\sqrt{\frac{1+x}{1-x}}}{dx}=\frac{\sqrt{1-x}}{2\sqrt{1+x}}\frac{2}{(1-x)^2}=\frac1{(1-x)\sqrt{1-x^2}}$$

So the antiderivative is, $$e^x\sqrt{\frac{1+x}{1-x}}+C$$

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    $\begingroup$ thanks, I just spotted a mistake of $-1$ in my calculations-that doesn't leave a $x$ in the numerator. Thanks for your toil! $\endgroup$ – Ashish Gaurav Feb 23 '13 at 14:07

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