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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

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