What is the integral of (sqrt(y^2-25))/y ?

The integral of (sqrt(y^2-25))/y is -5arcsec(1/5 y)+sqrt(y^2-25)+C

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integral of (sqrt(y^2-25))/y

\int \frac{y ^{2} - 25}{y} d y

- 5 arcsec (\frac{1}{5} y) + y^{2} - 25 + C

Solution steps

\int \frac{y ^{2} - 25}{y} d y

Apply Trigonometric Substitution

: \int 5 tan^{2} (u) d u

= \int 5 tan^{2} (u) d u

Take the constant out:

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

= 5 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 5 \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

= 5 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 5 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{5} y)

= 5 (- arcsec (\frac{1}{5} y) + tan (arcsec (\frac{1}{5} y)))

Simplify

5 (- arcsec (\frac{1}{5} y) + tan (arcsec (\frac{1}{5} y))) : - 5 arcsec (\frac{1}{5} y) + y^{2} - 25

= - 5 arcsec (\frac{1}{5} y) + y^{2} - 25

Add a constant to the solution

= - 5 arcsec (\frac{1}{5} y) + y^{2} - 25 + C

x \to 2 - lim (\frac{1}{x ^{3} - 2 x ^{2}})

\int x cos^{2} (2 x) d x

\frac{( - 1 )}{s ^{2} + 5 s + 6}

f (x) = \frac{3 x - 1}{x}

\int \frac{cos ^{5} ( x )}{sin ^{3} ( x )} d x

What is the integral of (sqrt(y^2-25))/y ?
The integral of (sqrt(y^2-25))/y is -5arcsec(1/5 y)+sqrt(y^2-25)+C