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

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

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integral of (sqrt(y^2-9))/y

\int \frac{y ^{2} - 9}{y} d y

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

Solution steps

\int \frac{y ^{2} - 9}{y} d y

Apply Trigonometric Substitution

: \int 3 tan^{2} (u) d u

= \int 3 tan^{2} (u) d u

Take the constant out:

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

= 3 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 3 \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

= 3 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 3 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{3} y)

= 3 (- arcsec (\frac{1}{3} y) + tan (arcsec (\frac{1}{3} y)))

Simplify

3 (- arcsec (\frac{1}{3} y) + tan (arcsec (\frac{1}{3} y))) : - 3 arcsec (\frac{1}{3} y) + y^{2} - 9

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

Add a constant to the solution

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

9 x - x^{2}, 2 x, 0, 7

\frac{d}{d x} (4 - 4 x^{2})

\frac{d ^{2} y}{d x ^{2}} + (\frac{d y}{d x})^{2} = 0

h (x) = x^{\frac{6}{5}}

\frac{d ^{2}}{d x ^{2}} (4 x^{3} - 3 x^{4})

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