Question

integrate 1 / x^3 sqrt(x^2 - 4)?

Answer 1

\[\int\frac{dx}{x^3\sqrt{x^2-4}}\]
Substitute: \(x=2\sec u\), then \(dx=2\sec u\tan u~du\):
\[\int\frac{2\sec u\tan u}{(2\sec u)^3\sqrt{(2\sec u)^2-4}}~du\\
\int\frac{2\sec u\tan u}{8\sec^3 u\sqrt{4\sec^2u-4}}~du\\
\frac{2}{8\sqrt4}\int\frac{\sec u\tan u}{\sec^3 u\sqrt{\sec^2u-1}}~du\\
\frac{1}{8}\int\frac{\sec u\tan u}{\sec^3 u\sqrt{\tan^2u}}~du\\
\frac{1}{8}\int\frac{\sec u\tan u}{\sec^3 u\tan u}~du\\
\frac{1}{8}\int\frac{du}{\sec^2 u}~du\\
\frac{1}{8}\int\cos^2u~du\]
Use this identity: \(\cos^2u=\dfrac{1}{2}(1+\cos2u)\):
\[\frac{1}{8}\int\frac{1}{2}(1+\cos2u)~du\\
\frac{1}{16}\int(1+\cos2u)~du\\
~~~~~~~~~~~\vdots\]