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

The integral of (y^2-b^2)/(y^2) is y+(b^2)/y+C

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integral of (y^2-b^2)/(y^2)

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

y + \frac{b ^{2}}{y} + C

Solution steps

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

Expand

\frac{y ^{2} - b ^{2}}{y ^{2}} : 1 - \frac{b ^{2}}{y ^{2}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d y = y

\int \frac{b ^{2}}{y ^{2}} d y = - \frac{b ^{2}}{y}

= y - (- \frac{b ^{2}}{y})

Simplify

= y + \frac{b ^{2}}{y}

Add a constant to the solution

= y + \frac{b ^{2}}{y} + C

\int \frac{e ^{x}}{1 + e ^{x}} d x

x \to 8 lim (\frac{( 1 )}{( x - 8 )})

\frac{\partial}{\partial y} (arcsin (x y))

\int \frac{1}{x ^{2} - 4 x + 9} d x

f (x) = 6 x - x^{2}, at x = 1

What is the integral of (y^2-b^2)/(y^2) ?
The integral of (y^2-b^2)/(y^2) is y+(b^2)/y+C