What is the integral of (sqrt(y^2-49))/y ?

The integral of (sqrt(y^2-49))/y is -7arcsec(1/7 y)+sqrt(y^2-49)+C

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integral of (sqrt(y^2-49))/y

\int \frac{y ^{2} - 49}{y} d y

- 7 arcsec (\frac{1}{7} y) + y^{2} - 49 + C

Solution steps

\int \frac{y ^{2} - 49}{y} d y

Apply Trigonometric Substitution

: \int 7 tan^{2} (u) d u

= \int 7 tan^{2} (u) d u

Take the constant out:

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

= 7 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 7 \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

= 7 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 7 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{7} y)

= 7 (- arcsec (\frac{1}{7} y) + tan (arcsec (\frac{1}{7} y)))

Simplify

7 (- arcsec (\frac{1}{7} y) + tan (arcsec (\frac{1}{7} y))) : - 7 arcsec (\frac{1}{7} y) + y^{2} - 49

= - 7 arcsec (\frac{1}{7} y) + y^{2} - 49

Add a constant to the solution

= - 7 arcsec (\frac{1}{7} y) + y^{2} - 49 + C

\int sin (12 x) cos (5 x) d x

\int_{- \infty}^{0} \frac{1}{5 - 4 x} d x

\frac{d}{d x} (ln (13 x) + 13)

\frac{d y}{d x} = \frac{y ^{2} + 3 x x ^{2} + y ^{2}}{x y}

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

What is the integral of (sqrt(y^2-49))/y ?
The integral of (sqrt(y^2-49))/y is -7arcsec(1/7 y)+sqrt(y^2-49)+C