What is the integral of (sqrt(y^2-144))/y ?

The integral of (sqrt(y^2-144))/y is -12arcsec(1/12 y)+sqrt(y^2-144)+C

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integral of (sqrt(y^2-144))/y

\int \frac{y ^{2} - 144}{y} d y

- 12 arcsec (\frac{1}{12} y) + y^{2} - 144 + C

Solution steps

\int \frac{y ^{2} - 144}{y} d y

Apply Trigonometric Substitution

: \int 12 tan^{2} (u) d u

= \int 12 tan^{2} (u) d u

Take the constant out:

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

= 12 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 12 \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

= 12 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 12 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{12} y)

= 12 (- arcsec (\frac{1}{12} y) + tan (arcsec (\frac{1}{12} y)))

Simplify

12 (- arcsec (\frac{1}{12} y) + tan (arcsec (\frac{1}{12} y))) : - 12 arcsec (\frac{1}{12} y) + y^{2} - 144

= - 12 arcsec (\frac{1}{12} y) + y^{2} - 144

Add a constant to the solution

= - 12 arcsec (\frac{1}{12} y) + y^{2} - 144 + C

f (x) = x^{5} cos (x)

\int \frac{1}{x ( x ^{2} + x + 5 )} d x

\frac{\partial}{\partial x} ((6 - x^{2} y^{2})^{2})

x \to 4 + lim (\frac{4 + x}{4 - x})

h \to 0 lim (6 x + 1 + 3 h)

What is the integral of (sqrt(y^2-144))/y ?
The integral of (sqrt(y^2-144))/y is -12arcsec(1/12 y)+sqrt(y^2-144)+C