What is the integral of (ln(3x))/(4x) ?

The integral of (ln(3x))/(4x) is 1/8 ln^2(3x)+C

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integral of (ln(3x))/(4x)

\int \frac{ln ( 3 x )}{4 x} d x

\frac{1}{8} ln^{2} (3 x) + C

Solution steps

\int \frac{ln ( 3 x )}{4 x} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{ln ( 3 x )}{x} d x

Apply u-substitution

= \frac{1}{4} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

= \frac{1}{4} \cdot \int v d v

Apply the Power Rule

= \frac{1}{4} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{4} \cdot \frac{ln ^{2} ( 3 x )}{2}

Simplify

\frac{1}{4} \cdot \frac{ln ^{2} ( 3 x )}{2} : \frac{1}{8} ln^{2} (3 x)

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

Add a constant to the solution

= \frac{1}{8} ln^{2} (3 x) + C

What is the integral of (ln(3x))/(4x) ?
The integral of (ln(3x))/(4x) is 1/8 ln^2(3x)+C