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

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

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

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

- b arctan (\frac{y ^{2} - b ^{2}}{b}) + y^{2} - b^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} + b ^{2}} d u

= \int \frac{u ^{2}}{u ^{2} + b ^{2}} d u

Apply Integral Substitution

= \int \frac{b v ^{2}}{v ^{2} + 1} d v

Take the constant out:

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

= b \cdot \int \frac{v ^{2}}{v ^{2} + 1} d v

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

= b \cdot \int - \frac{1}{v ^{2} + 1} + 1 d v

Apply the Sum Rule:

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

= b (- \int \frac{1}{v ^{2} + 1} d v + \int 1 d v)

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= b (- arctan (v) + v)

Substitute back

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

Simplify

b (- arctan (\frac{y ^{2} - b ^{2}}{b}) + \frac{y ^{2} - b ^{2}}{b}) : - b arctan (\frac{y ^{2} - b ^{2}}{b}) + y^{2} - b^{2}

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

Add a constant to the solution

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

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