What is the integral of (sqrt(y^2-100))/y ?

The integral of (sqrt(y^2-100))/y is -10arcsec(1/10 y)+sqrt(y^2-100)+C

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integral of (sqrt(y^2-100))/y

\int \frac{y ^{2} - 100}{y} d y

- 10 arcsec (\frac{1}{10} y) + y^{2} - 100 + C

Solution steps

\int \frac{y ^{2} - 100}{y} d y

Apply Trigonometric Substitution

: \int 10 tan^{2} (u) d u

= \int 10 tan^{2} (u) d u

Take the constant out:

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

= 10 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 10 \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

= 10 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 10 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{10} y)

= 10 (- arcsec (\frac{1}{10} y) + tan (arcsec (\frac{1}{10} y)))

Simplify

10 (- arcsec (\frac{1}{10} y) + tan (arcsec (\frac{1}{10} y))) : - 10 arcsec (\frac{1}{10} y) + y^{2} - 100

= - 10 arcsec (\frac{1}{10} y) + y^{2} - 100

Add a constant to the solution

= - 10 arcsec (\frac{1}{10} y) + y^{2} - 100 + C

\int \frac{24 x ^{3} - 48 x ^{2} + 5}{x ^{2} - 2 x} d x

\frac{1}{4 x + 3}

x - x^{2} - x^{2} + 6 x, 0, 3.5

y = x^{2}, y = 0, x = 1

\int \frac{w ^{2} - 25}{w ^{3}} d w

What is the integral of (sqrt(y^2-100))/y ?
The integral of (sqrt(y^2-100))/y is -10arcsec(1/10 y)+sqrt(y^2-100)+C