What is the integral of (sqrt(y^2-4))/y ?

The integral of (sqrt(y^2-4))/y is -2arcsec(1/2 y)+sqrt(y^2-4)+C

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integral of (sqrt(y^2-4))/y

\int \frac{y ^{2} - 4}{y} d y

- 2 arcsec (\frac{1}{2} y) + y^{2} - 4 + C

Solution steps

\int \frac{y ^{2} - 4}{y} d y

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= 2 \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

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

\int 1 d u = u

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

= 2 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{2} y)

= 2 (- arcsec (\frac{1}{2} y) + tan (arcsec (\frac{1}{2} y)))

Simplify

2 (- arcsec (\frac{1}{2} y) + tan (arcsec (\frac{1}{2} y))) : - 2 arcsec (\frac{1}{2} y) + y^{2} - 4

= - 2 arcsec (\frac{1}{2} y) + y^{2} - 4

Add a constant to the solution

= - 2 arcsec (\frac{1}{2} y) + y^{2} - 4 + C

\frac{d}{d x} (\frac{x ^{5} - 8 x ^{3} + x + 6}{x})

x \to 0 lim (2 x^{3} - 4 x + 2)

\int (\frac{2}{x} + 3 sin (x) - 4 e^{x}) d x

y = 1 + 3 x, 1 \leq x \leq 9

x \to \infty lim (a^{x})

What is the integral of (sqrt(y^2-4))/y ?
The integral of (sqrt(y^2-4))/y is -2arcsec(1/2 y)+sqrt(y^2-4)+C