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

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

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

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

\frac{1}{3} (ln ∣ y ∣ - \frac{y ^{2}}{2}) + C

Solution steps

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

Take the constant out:

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

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

Expand

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

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

\int \frac{1}{y} d y = ln ∣ y ∣

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

= \frac{1}{3} (ln ∣ y ∣ - \frac{y ^{2}}{2})

Add a constant to the solution

= \frac{1}{3} (ln ∣ y ∣ - \frac{y ^{2}}{2}) + C

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