Find $\int \frac{dx}{x\sqrt{x^2-2}}$ $\int \frac{dx}{x\sqrt{x^2-2}}$
My book says to solve this using $x = 1/t$ so I did $dx = -1/t^2$
$$-\int\frac{\frac{1}{t^2}}{t\sqrt{1-2t^2}} = -\int \frac{1}{\sqrt{1-2t^2}} = -\arcsin(\sqrt{2t})=-\arcsin(\sqrt{2}/x)$$
My book says the solution is $\frac{1}{\sqrt{2}}\arccos(\frac{\sqrt{2}}{x}).$ What went wrong?

First, remember integration constants: $\arcsin y=\frac{\pi}{2}-\arccos y$. Secondly, be careful with coefficients: $\sqrt{1/t^2-2}=\sqrt{2}\sqrt{1/(2t^2)-1}$.
