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

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

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integral of (ln(y^2))/(3y)

\int \frac{ln ( y ^{2} )}{3 y} d y

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

Solution steps

\int \frac{ln ( 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{ln ( y ^{2} )}{y} d y

Apply u-substitution

: \int \frac{ln ( u )}{2 u} d u

= \frac{1}{3} \cdot \int \frac{ln ( u )}{2 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{ln ^{2} ( y ^{2} )}{2} : \frac{1}{3} ln^{2} (y)

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

Add a constant to the solution

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

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