What is the integral of ((sqrt(x^2-9)))/(x^3) ?

The integral of ((sqrt(x^2-9)))/(x^3) is -(sqrt(x^2-9))/(2x^2)+1/6 arcsec(1/3 x)+C

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integral of ((sqrt(x^2-9)))/(x^3)

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

- \frac{x ^{2} - 9}{2 x ^{2}} + \frac{1}{6} arcsec (\frac{1}{3} x) + C

Solution steps

\int \frac{x ^{2} - 9}{x ^{3}} d x

Apply Integration By Parts

: - \frac{x ^{2} - 9}{2 x ^{2}} - \int - \frac{1}{2 x x ^{2} - 9} d x

= - \frac{x ^{2} - 9}{2 x ^{2}} - \int - \frac{1}{2 x x ^{2} - 9} d x

\int - \frac{1}{2 x x ^{2} - 9} d x = - \frac{1}{6} arcsec (\frac{1}{3} x)

= - \frac{x ^{2} - 9}{2 x ^{2}} - (- \frac{1}{6} arcsec (\frac{1}{3} x))

Simplify

= - \frac{x ^{2} - 9}{2 x ^{2}} + \frac{1}{6} arcsec (\frac{1}{3} x)

Add a constant to the solution

= - \frac{x ^{2} - 9}{2 x ^{2}} + \frac{1}{6} arcsec (\frac{1}{3} x) + C

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What is the integral of ((sqrt(x^2-9)))/(x^3) ?
The integral of ((sqrt(x^2-9)))/(x^3) is -(sqrt(x^2-9))/(2x^2)+1/6 arcsec(1/3 x)+C