What is the integral of (sqrt(y^2-25))/(y^3) ?

The integral of (sqrt(y^2-25))/(y^3) is -(sqrt(y^2-25))/(2y^2)+1/10 arcsec(1/5 y)+C

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integral of (sqrt(y^2-25))/(y^3)

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

- \frac{y ^{2} - 25}{2 y ^{2}} + \frac{1}{10} arcsec (\frac{1}{5} y) + C

Solution steps

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

Apply Integration By Parts

: - \frac{y ^{2} - 25}{2 y ^{2}} - \int - \frac{1}{2 y y ^{2} - 25} d y

= - \frac{y ^{2} - 25}{2 y ^{2}} - \int - \frac{1}{2 y y ^{2} - 25} d y

\int - \frac{1}{2 y y ^{2} - 25} d y = - \frac{1}{10} arcsec (\frac{1}{5} y)

= - \frac{y ^{2} - 25}{2 y ^{2}} - (- \frac{1}{10} arcsec (\frac{1}{5} y))

Simplify

= - \frac{y ^{2} - 25}{2 y ^{2}} + \frac{1}{10} arcsec (\frac{1}{5} y)

Add a constant to the solution

= - \frac{y ^{2} - 25}{2 y ^{2}} + \frac{1}{10} arcsec (\frac{1}{5} y) + C

\int \frac{1}{600 - x} d x

\frac{d y}{d x}, 3 x - tan (y) = 4

\int_{3}^{5} \frac{2 t}{t ^{2} - 4 t + 4} d t

y^{^{'}} = \frac{4 x ^{3} + y ^{3}}{x y ^{2}}

What is the integral of (sqrt(y^2-25))/(y^3) ?
The integral of (sqrt(y^2-25))/(y^3) is -(sqrt(y^2-25))/(2y^2)+1/10 arcsec(1/5 y)+C