What is the integral of 1/x ln(3x) ?

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

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integral of 1/x ln(3x)

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

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

Solution steps

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

Simplify

\frac{1}{x} ln (3 x) : \frac{ln ( 3 x )}{x}

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

Apply u-substitution

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

= \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \int v d v

Apply the Power Rule

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

= \frac{v ^{2}}{2}

Substitute back

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

Add a constant to the solution

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

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