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I am trying to find the indefinite integral of $\frac{1}{\sqrt{3x^2-1}}$ using the table of basic indefinite integrals (no substitution or integration by parts). I considered using the formula primitive of $\frac{1}{\sqrt{a^2-x^2}} = \arcsin\bigl(\frac x a\bigr) + K$ with $a$ greater than $0$ for this case but the a is negative and there is a $3$ multiplying the $x^2$. Can someone help me solve this?

Thank you!

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It seems like based on the results, trig functions (and their inverses) are not going to suit you well. This requires the knowledge of Hyperbolic Functions (Hyperbolic Sine, Hyperbolic Cosine etc) and their inverses. As a general rule, we find that for the inverse Hyperbolic Cosine: $$\frac{d}{dx}\cosh^{-1}ax=\frac{a}{\sqrt{(ax)^2-1}}$$ Thus, the solution is given as $$\int\frac{1}{\sqrt{3x^2-1}}dx=\frac{1}{\sqrt{3}}\cosh^{-1}(\sqrt{3} x)$$

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$\frac{1}{\sqrt{3x^2-1}}=\frac{1}{\sqrt3} \frac{1}{\sqrt{x^2-\frac{1}{3}}} $ and you need formula $\frac{1}{\sqrt{x^2-a^2}} = \log(x+\sqrt{x^2-a^2}) + K$

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