What is the integral of (y^2-1)/(y 3/2) ?

The integral of (y^2-1)/(y 3/2) is (y^2)/3-2/3 ln|y|+C

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integral of (y^2-1)/(y 3/2)

\int \frac{y ^{2} - 1}{y \frac{3}{2}} d y

\frac{y ^{2}}{3} - \frac{2}{3} ln ∣ y ∣ + C

Solution steps

\int \frac{y ^{2} - 1}{y \frac{3}{2}} d y

Take the constant out:

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

= \frac{1}{\frac{3}{2}} \cdot \int \frac{y ^{2} - 1}{y} d y

Simplify

= \frac{2}{3} \cdot \int \frac{y ^{2} - 1}{y} d y

Expand

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

= \frac{2}{3} \cdot \int y - \frac{1}{y} 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

= \frac{2}{3} (\int y d y - \int \frac{1}{y} d y)

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

\int \frac{1}{y} d y = ln ∣ y ∣

= \frac{2}{3} (\frac{y ^{2}}{2} - ln ∣ y ∣)

Simplify

\frac{2}{3} (\frac{y ^{2}}{2} - ln ∣ y ∣) : \frac{y ^{2}}{3} - \frac{2}{3} ln ∣ y ∣

= \frac{y ^{2}}{3} - \frac{2}{3} ln ∣ y ∣

Add a constant to the solution

= \frac{y ^{2}}{3} - \frac{2}{3} ln ∣ y ∣ + C

What is the integral of (y^2-1)/(y 3/2) ?
The integral of (y^2-1)/(y 3/2) is (y^2)/3-2/3 ln|y|+C