Question

find the integral of 1/sqrt(x^2-9)

Answer 1

Let \(x=3\sec u\), so \(dx=3\sec u\tan u~du\):
\[\int\frac{dx}{\sqrt{x^2-9}}=3\int\frac{\sec u\tan u}{\sqrt{9\sec^2u-9}}~du=\int\frac{\sec u\tan u}{\sqrt{\sec^2u-1}}~du\]