<s>[INST] How to integrate $\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$? How to integrate $$\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$$
I tried both $t=\sqrt{x^2-1}$ and $t=\sin x$ but didn't reach the right result. 
 [/INST] Apply the substitution $x=sec(u)$, then $dx=\frac{\tan \left(u\right)}{\cos \left(u\right)}du$.
$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{\tan \left(u\right)}{\cos \left(u\right)}du$$
$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{1}{\cos^2 \left(u\right)}du$$
Remember that $\frac {1}{\cos(u)}=\sec(u)$ :
$$\int \frac{\sin \left(u\right)}{\sqrt{\sec ^2\left(u\right)-1}}du$$
And $\sec ^2\left(x\right)=1+\tan ^2\left(x\right)$:
$$\int \frac{\sin \left(u\right)}{\sqrt{-1+1+\tan ^2\left(u\right)}}du$$
Then:
$$\int \frac{\sin \left(u\right)}{\sqrt{\tan ^2\left(u\right)}}du$$
$$if \int \frac{g\left(x\right)}{f\left(x\right)}=F\left(x\right)\mathrm{,\:then\:}\:\int \frac{g\left(x\right)}{\sqrt{f^2\left(x\right)}}=\frac{f\left(x\right)}{\sqrt{f^2\left(x\right)}}F\left(x\right)$$
$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\int \frac{\sin \left(u\right)}{\tan \left(u\right)}du$$
$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\int \cos \left(u\right)du$$
$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\sin \left(u\right)$$
Substitute back, $u=arcsec(u)$:
$$\frac{\tan \left(arcsec \left(x\right)\right)}{\sqrt{\left(\tan \left(arcsec \left(x\right)\right)\right)^2}}\sin \left(arcsec \left(x\right)\right)$$
Simplifies to:
$$\sqrt{1-\frac{1}{x^2}}+C$$
Now compute the boundaries and you will find $1-0=1$.
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