What is the integral of (sqrt(x^2-81))/x ?

The integral of (sqrt(x^2-81))/x is -9arcsec(1/9 x)+sqrt(x^2-81)+C

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integral of (sqrt(x^2-81))/x

\int \frac{x ^{2} - 81}{x} d x

- 9 arcsec (\frac{1}{9} x) + x^{2} - 81 + C

Solution steps

\int \frac{x ^{2} - 81}{x} d x

Apply Trigonometric Substitution

: \int 9 tan^{2} (u) d u

= \int 9 tan^{2} (u) d u

Take the constant out:

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

= 9 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 9 \cdot \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= 9 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{9} x)

= 9 (- arcsec (\frac{1}{9} x) + tan (arcsec (\frac{1}{9} x)))

Simplify

9 (- arcsec (\frac{1}{9} x) + tan (arcsec (\frac{1}{9} x))) : - 9 arcsec (\frac{1}{9} x) + x^{2} - 81

= - 9 arcsec (\frac{1}{9} x) + x^{2} - 81

Add a constant to the solution

= - 9 arcsec (\frac{1}{9} x) + x^{2} - 81 + C

\int \frac{( x ^{2} - 1 )}{( x ^{2} + 1 ) ( x - 2 )} d x

\frac{\partial}{\partial x} (\frac{5 x - 3 y}{5 x + 3 y})

\int x^{2} cos (x^{3}) d x

x \to 1 lim (- x^{2} + 9 x - 7)

(1 + e^{x}) \frac{d y}{d x} + e^{x} y = 0

What is the integral of (sqrt(x^2-81))/x ?
The integral of (sqrt(x^2-81))/x is -9arcsec(1/9 x)+sqrt(x^2-81)+C