Hey, I have an integral: $$\int_{1}^{\infty}\dfrac{dx}{x\sqrt{x^2 - 1}}$$. My answer is, how to solve it.

Seeing that square in a square root, your first thought should be "trig substitution". $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$ so $\displaystyle cos^2(\theta)- 1= -sin^2(\theta)$ , dividing both sides by $\displaystyle cos^2(\theta)$, $\displaystyle 1- sec^2(\theta)= tan^2(\theta)$. Let $\displaystyle x= sec(\theta)$ so that $\displaystyle dx= sec(\theta)tan(\theta)d\theta$. The integral becomes $\displaystyle \int\frac{sec( \theta )tan( \theta )d \theta}{sec(\theta)tan(\theta)}$. That should be easy to integrate! (In fact it is a standard integral on "tables of integrals.)

Hey, I have an integral: $$\int_{1}^{\infty}\dfrac{dx}{x\sqrt{x^2 - 1}}$$. My answer is, how to solve it.
$\displaystyle \int {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} = \arctan \left( {\frac{1}{{\sqrt {{x^2} - 1} }}} \right)$
$\displaystyle \mathop {\lim }\limits_{x \to \infty } \arctan \left( {\frac{1}{{\sqrt {{x^2} - 1} }}} \right)=~?$______&_____$\displaystyle \mathop {\lim }\limits_{x \to {1^ + }} \arctan \left( {\frac{1}{{\sqrt {{x^2} - 1} }}} \right)=~?$
Yes. First limit is $\displaystyle \dfrac{\pi}{2}$, the second limit is $\displaystyle 0$.