What is the integral of (sqrt(y^2-64))/y ?

The integral of (sqrt(y^2-64))/y is -8arcsec(1/8 y)+sqrt(y^2-64)+C

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integral of (sqrt(y^2-64))/y

\int \frac{y ^{2} - 64}{y} d y

- 8 arcsec (\frac{1}{8} y) + y^{2} - 64 + C

Solution steps

\int \frac{y ^{2} - 64}{y} d y

Apply Trigonometric Substitution

: \int 8 tan^{2} (u) d u

= \int 8 tan^{2} (u) d u

Take the constant out:

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

= 8 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 8 \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

= 8 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 8 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{8} y)

= 8 (- arcsec (\frac{1}{8} y) + tan (arcsec (\frac{1}{8} y)))

Simplify

8 (- arcsec (\frac{1}{8} y) + tan (arcsec (\frac{1}{8} y))) : - 8 arcsec (\frac{1}{8} y) + y^{2} - 64

= - 8 arcsec (\frac{1}{8} y) + y^{2} - 64

Add a constant to the solution

= - 8 arcsec (\frac{1}{8} y) + y^{2} - 64 + C

\int z e^{y} + x^{6} d y

\frac{d}{d x} (sin (x) sin (y))

3 (\frac{d y}{d x}) + 12 y = 4

\int \frac{sin ^{3} ( x )}{x} d x

\frac{\partial}{\partial x} (x ln (z^{4} + x^{6}))

What is the integral of (sqrt(y^2-64))/y ?
The integral of (sqrt(y^2-64))/y is -8arcsec(1/8 y)+sqrt(y^2-64)+C