Question

integrate 1/sqrt(x^2-1) from 2 to 3?

Answer 1

you have a form \(\sqrt{x^2-a^2}\) suggesting to use the trigonometric substitution
\(x = \sec \theta\)
\(dx = \sec \theta \tan \theta \, d\theta\)

Ignoring the bounds for now,
\[ \int \frac{1}{\sqrt{x^2-1}}dx=\int \frac{1}{\sqrt{\sec^2 \theta-1}}(\sec \theta \tan \theta \, d\theta)\\
\, \\
= \int\frac{\sec \theta \tan \theta}{\tan \theta}d\theta, \text{ since } \sec^2\theta -1 = \tan^2 \theta \\
\, \\
= \int \sec \theta d \theta \\
= \ln|\sec \theta +\tan \theta |+C\]

Now into x-variables again, notice that \(x = \sec \theta\) and \(\sqrt{x^2-1}=\tan \theta\) from the simplification in your denominator, so
\[=\ln\left|x+\sqrt{x^2-1}\right|+C \]
You can drop the absolute value bars since you will be integrating from 2 to 3.
With the bounds:
\[ \large = \left [ \ln \left(x+\sqrt{x^2-1}\right) \right]_{2}^3 = \ln(3+\sqrt{3^2-1})-\ln(2+\sqrt{2^2-1}) \\
\large =\ln(3+2\sqrt{2})-\ln(2+\sqrt{3})\]