Question

Evaluate the integral \(\int {\frac{{\sqrt {{x^2} - 9} }}{{{x^3}}}dx} \)

Short answer

The evaluation of the equation gives us, \(\int {\frac{{\sqrt {{x^2} - 9} }}{{{x^3}}}dx} = \frac{1}{6}se{c^{ - 1}}\left( {\frac{x}{3}} \right) - \frac{1}{2}\frac{{\sqrt {{x^2} - 9} }}{x} + c\)

Step-by-step solution

Simplify the question

\(\int {\frac{{\sqrt {{x^2} - 9} }}{{{x^3}}}dx} \)

Let

\(\begin{array}{l}x = 3sec\theta \\dx = 3sec\theta tan\theta d\theta \\\theta = se{c^{ - 1}}\left( {\frac{x}{3}} \right)\\\int {\left( {\frac{{\sqrt {{x^2} - 9} }}{{{x^3}}}} \right)dx = \int {\frac{{\sqrt {9se{c^2}\theta - 9} }}{{27se{c^3}\theta }}3sec\theta tan\theta d\theta } } \\ = \frac{1}{9}\int {\frac{{\sqrt {9\left( {se{c^2}\theta - 1} \right)} }}{{se{c^2}\theta }}} tan\theta d\theta \end{array}\)

Use \(se{c^2}\theta - 1 = ta{n^2}\theta \& \sqrt 9 = 3\)

\(\begin{aligned}{l} &= \frac{3}{9}\int {\frac{{\sqrt {ta{n^2}\theta } }}{{se{c^2}\theta }}tan\theta d\theta } \\ &= \frac{1}{3}\int {\frac{{tan\theta }}{{se{c^2}\theta }}tan\theta d\theta } \\ &= \frac{1}{3}\int {\frac{{ta{n^2}\theta }}{{se{c^2}\theta }}d\theta } \end{aligned}\)

Change \(\theta \) to \(x\) for final answer.

\(\int {\frac{{\sqrt {{x^2} - 9} }}{{{x^3}}}dx} = \frac{1}{6}se{c^{ - 1}}\left( {\frac{x}{3}} \right) - \frac{1}{6} \times \frac{1}{2} \times 2sin\theta cos\theta \)

We have \(sec\theta = \frac{x}{3},sin\theta = \frac{{\sqrt {{x^2} - 9} }}{x},cos\theta = \frac{3}{x}\)

\(\begin{aligned}{l} &= \frac{1}{6}se{c^{ - 1}}\left( {\frac{x}{3}} \right) - \frac{1}{6}\frac{{\sqrt {{x^2} - 9} }}{x}\frac{3}{x}\\ &= \frac{1}{6}se{c^{ - 1}}\left( {\frac{x}{3}} \right) - \frac{1}{2}\frac{{\sqrt {{x^2} - 9} }}{x} + c\end{aligned}\)

Hence, \(\int {\frac{{\sqrt {{x^2} - 9} }}{{{x^3}}}dx} = \frac{1}{6}se{c^{ - 1}}\left( {\frac{x}{3}} \right) - \frac{1}{2}\frac{{\sqrt {{x^2} - 9} }}{x} + c\)