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

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

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

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

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

Solution steps

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

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} y)

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

Simplify

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

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

Add a constant to the solution

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

n = 0 \sum \infty \frac{1}{2 ^{n} + 3 ^{n}}

\int cos (2 n x) d x

9 (t + 1) \frac{d y}{d t} - 7 y = 14 t

\frac{\partial}{\partial x} (- 5 x^{2} + 7 y^{5} x)

h (x) = tan (4 x^{2} - 3 x - 4)

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