Answer
\[\frac{-1}{2}\sin^{-1}\left(\frac{2}{x}\right)\]
Work Step by Step
Let \[I=\int\frac{dx}{x\sqrt {x^2-4}}\]
Put \[x=\frac{1}{t}\;\;\Rightarrow \;\;dx=\frac{-1}{t^2}dt\]
\[I=\int \frac{\frac{-1}{t^2}}{\frac{1}{t}\sqrt{\frac{1}{t^2}-4}}dt\]
\[\Rightarrow I=-\int\frac{dt}{\sqrt{1-4t^2}}\]
\[I=-\frac{1}{2}\int\frac{dt}{\sqrt{\left(\frac{1}{2}\right)^2-t^2}}\]
\[\Rightarrow I=\frac{-1}{2}\sin^{-1}(2t)\]
\[\Rightarrow I=\frac{-1}{2}\sin^{-1}\left(\frac{2}{x}\right)\]
Hence \[I=\frac{-1}{2}\sin^{-1}\left(\frac{2}{x}\right)\]
