What is the integral of (ln(y^3))/(8y) ?

The integral of (ln(y^3))/(8y) is 3/16 ln^2(y)+C

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integral of (ln(y^3))/(8y)

\int \frac{ln ( y ^{3} )}{8 y} d y

\frac{3}{16} ln^{2} (y) + C

Solution steps

\int \frac{ln ( y ^{3} )}{8 y} d y

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{ln ( y ^{3} )}{y} d y

Apply u-substitution

: \int \frac{ln ( u )}{3 u} d u

= \frac{1}{8} \cdot \int \frac{ln ( u )}{3 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \frac{1}{3} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \frac{1}{8} \cdot \frac{1}{3} \cdot \int v d v

Apply the Power Rule

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

= \frac{1}{8} \cdot \frac{1}{3} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{8} \cdot \frac{1}{3} \cdot \frac{ln ^{2} ( y ^{3} )}{2}

Simplify

\frac{1}{8} \cdot \frac{1}{3} \cdot \frac{ln ^{2} ( y ^{3} )}{2} : \frac{3}{16} ln^{2} (y)

= \frac{3}{16} ln^{2} (y)

Add a constant to the solution

= \frac{3}{16} ln^{2} (y) + C

f (x) = x^{2} + 3 x - 1, at x = - 1

\int \frac{1}{x 9 x ^{2} - 49} d x

\frac{d}{d x} (\frac{1}{( 1 + x ) ^{3}})

4 y^{^{''}} + 16 y^{^{'}} + 19 y = 0

\frac{d}{d x} (arctan (\frac{d}{x}))

What is the integral of (ln(y^3))/(8y) ?
The integral of (ln(y^3))/(8y) is 3/16 ln^2(y)+C