What is the integral of (sqrt(y^2-16))/y ?

The integral of (sqrt(y^2-16))/y is -4arcsec(1/4 y)+sqrt(y^2-16)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(y^2-16))/y

\int \frac{y ^{2} - 16}{y} d y

- 4 arcsec (\frac{1}{4} y) + y^{2} - 16 + C

Solution steps

\int \frac{y ^{2} - 16}{y} d y

Apply Trigonometric Substitution

: \int 4 tan^{2} (u) d u

= \int 4 tan^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 4 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{4} y)

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

Simplify

4 (- arcsec (\frac{1}{4} y) + tan (arcsec (\frac{1}{4} y))) : - 4 arcsec (\frac{1}{4} y) + y^{2} - 16

= - 4 arcsec (\frac{1}{4} y) + y^{2} - 16

Add a constant to the solution

= - 4 arcsec (\frac{1}{4} y) + y^{2} - 16 + C

f (x) = ln (1 - x^{2})

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

n = 1 \sum \infty 2 0^{n} x^{n} n!

\int_{- π}^{2 π} sin^{2} (x) d x

e^{6}

What is the integral of (sqrt(y^2-16))/y ?
The integral of (sqrt(y^2-16))/y is -4arcsec(1/4 y)+sqrt(y^2-16)+C