# Integrals and principle root question

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?

• $$\int \frac {dx} {x^2 \sqrt {x^2 + 1}} = -\sqrt {1 + x^{-2}}$$ 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. – 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.
• 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? – LHC2012 Jun 2 '19 at 8:41