My question is about the validity of taking out $x^2$ from principle roots during evaluation of certain integrals. For instance when solving: $$\int \frac{1}{x\cdot \sqrt{9x^2-1}}dx$$ Rather than directly subbing $u=\sec x$ I took the longer approach of doing: $$=\int \frac{1}{x\cdot \sqrt{x^2 \cdot \left(9-\frac{1}{x^2} \right)}}dx$$ $$=\int \frac{1}{x^2\cdot \sqrt{9-\frac{1}{x^2}}}dx$$ and then subbing $u=\frac{1}{x^2}$ to obtain: $$-\arcsin\left(\frac{1}{3x}\right)+C$$

Would this final answer be considered incorrect? How should I interpret how valid the result is? I'm thinking it is valid for positive $x$'s only?

Edit: In Blackpenredpen's $100$ integral videos, for the $22$th integral he also used this method, so why is it valid in his case?

  • $\begingroup$ $$\int \frac {dx} {x^2 \sqrt {x^2 + 1}} = -\sqrt {1 + x^{-2}}$$ is valid under certain assumptions. If you try to evaluate the integral over, say, $[-2, -1]$ as $F(-1) - F(-2)$, you'll get a negative value, while the integral is positive. $\endgroup$ – Maxim Jun 2 '19 at 12:47

Not quite. If you differentiate $\displaystyle-\arcsin\left(\frac1{3x}\right)$, then what you get is $\displaystyle\frac1{x^2\sqrt{9-\frac1{x^2}}}$, which is equal to $\displaystyle\frac1{x\sqrt{9x^2-1}}$ on $(0,\infty)$, but not in general.

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  • $\begingroup$ Thank you very much. What about the 22th integral in blackpenredpen's video which involves the same process for $\int \frac{1}{x^2 \cdot \sqrt{x^2+1}}dx$? Do you think it is considered valid and why? $\endgroup$ – LHC2012 Jun 2 '19 at 8:41
  • $\begingroup$ I do not know what that video is. Anyway, my goal was to answer your question and I suppose that I did that. $\endgroup$ – José Carlos Santos Jun 2 '19 at 8:43
  • $\begingroup$ Sorry for not clarifying, but the question I am talking about is in here: youtube.com/watch?v=dgm4-3-Iv3s at 1 hour and 16 minutes, where he used the same method $\endgroup$ – LHC2012 Jun 2 '19 at 9:20

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