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?

  • 2
    $\begingroup$ $f(x)$ 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

Your trick does work.


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, $$e^x\sqrt{\frac{1+x}{1-x}}+C$$

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.