How does one solve the following integral:
$$\int \frac{1}{\sqrt{\frac{C}{x^2}-1}}\;dx\;\;,$$
where $C$ is some constant. Should substitution be used here?
2026-04-11 20:20:04.1775938804
How to solve $\int \frac{1}{\sqrt{\frac{C}{x^2}-1}}\;dx\;\;$
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$$ \int \frac{1}{\sqrt{\frac{C}{x^2}-1}}dx = \int \frac{1}{\sqrt{C-x^2}}\frac{1}{\frac{1}{x}}dx=\\ \int \frac{xdx}{\sqrt{C-x^2}} $$ us the sub $C-x^2 = u$