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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \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

= 9 (- \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)

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

Simplify

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

Add a constant to the solution

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