Answer
$\dfrac{3}{2} \sec^{-1} x+C$
Work Step by Step
Calculate the anti-derivative.
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and
where $c$ is a constant of proportionality.
Then $\int \dfrac{3}{2x\sqrt {x^2-1}} dx=(\dfrac{3}{2})\int \dfrac{1}{x\sqrt {x^2-1}} dx$
As we know that $ \int \dfrac{3}{2x\sqrt {x^2-1}} dx=\sec^{-1} x+C$
Thus, $= \dfrac{3}{2} \sec^{-1} x+C$
