Integral of $\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$ I was asked to find the following integral: 
$$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
What I tried to replace $\sqrt{1-\frac{1}{x^2}}$ with $u$ so that: $$du=\frac{dx}{x^3\sqrt{1-\frac{1}{x^2}}} \Rightarrow du*x=\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
And:$$x=\sqrt{\frac{1}{1-u^2}}$$
And we can replace:
$$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}=\int\frac{du}{\sqrt{1-u^2}}=\arctan(u)+C=\arctan(\sqrt{1-\frac{1}{x^2}})+C$$
The problem is, when that result is derived we don't get the original expression. I just can't find my mistake, so some help would be appreciated.

HINT:
$$\dfrac1{x^2\sqrt{1-\dfrac1{x^2}}}=\dfrac{|x|}{x^2\sqrt{x^2-1}}$$
Set $\sqrt{x^2-1}=u\implies x^2=u^2+1$ and $x\ dx=u\ du$
