Evaluate the integral $\displaystyle \int \frac{1}{x \sqrt{x^2-4}} dx$

If we let $\displaystyle u = \frac{2}{x}$, then $\displaystyle x = \frac{2}{u}$ so...

$\displaystyle dx = \frac{-2}{u^2} du$

Thus,

$ \begin{equation} \begin{aligned} \int \frac{dx}{x \sqrt{x^2-4}} &= \int \frac{-2 du}{u^2 \left(\frac{2}{u}\right)\sqrt{\left(\frac{2}{u}\right)^2-4}}\\ \\ &= \int \frac{-2 du}{2u \sqrt{\frac{4}{u^2}-4}}\\ \\ &= \int \frac{-2 du}{2u \sqrt{\frac{4-4u^2}{u^2}}}\\ \\ &= \int \frac{-2 du}{2 \sqrt{4-4u^2}}\\ \\ &= -\int \frac{du}{\sqrt{4-4u^2}} = -\int \frac{du}{\sqrt{4(1-u^2)}} = -\int \frac{du}{2\sqrt{1-u^2}}\\ \\ \text{recall that } \frac{d}{dx} \sin^{-1} (x) &= \frac{1}{\sqrt{1-x^2}}\\ \\ &= -\frac{1}{2} \left[ \sin^{-1} u \right] + c \\ \\ &= -\frac{1}{2} \sin^{-1} \left( \frac{2}{x} \right) + c \end{aligned} \end{equation} $

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