Find the indefinite integral: $\int {1 \over {x^2 \sqrt {x^2-1}}}dx$ - more simple way?

Question

Find the indefinite integral: $$\int {1 \over {x^2 \sqrt {x^2-1}}}dx$$

I solved it using Integral Substitution where $t=\arccos{1 \over x}$. But is there a more simple way? (not $x = {1 \over \cos{x}}$, it's the same). Maybe without trigonometry?

The answer: $$\int {1 \over {x^2 \sqrt {x^2-1}}}dx = \sqrt{1- {1 \over {x^2}}}+C$$

Answer 1

Indeed, assuming $x>1$, factor one $x$ out from the square root, $$ \frac{1}{x^2\sqrt{x^2-1}}=\frac{1}{x^3\sqrt{1-1/x^2}}, $$ and let $u=1-1/x^2$. Everything will magically fall in place.

If $x<-1$, you can factor out $|x|=-x$ and do the same. You will get a similar result, but with a sign change. So, in fact, it might be better to write the primitive as $\sqrt{x^2-1}/x+C$. This is valid for all $|x|>1$.

Answer 2

$\int\frac{1}{x^2\sqrt{x^2-1}}dx=\int\frac{1}{x^3\sqrt{1-\frac{1}{x^2}}}dx$ $ = -\frac{1}{2}\int\frac{1}{\sqrt{1-\frac{1}{x^2}}}d\frac{1}{x^2}$ $ = \frac{1}{2}\int (1-\frac{1}{x^2})^{-1/2}d(1-\frac{1}{x^2})$ $ = (1-\frac{1}{x^2})^{1/2}+C$

Answer 3

$$\int {1 \over {x^2 \sqrt {x^2-1}}}dx=-\int{1 \over {\sqrt {x^2-1}}}d\left(\frac1x\right)=-\int\frac{\dfrac1x }{\sqrt{1-\dfrac1{x^2}}}d\left(\frac1x\right)=\sqrt{1-\frac1{x^2}}+C.$$