What is the integral of ln(2x) 1/x ?

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

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integral of ln(2x) 1/x

\int ln (2 x) \frac{1}{x} d x

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

Solution steps

\int ln (2 x) \frac{1}{x} d x

Simplify

ln (2 x) \frac{1}{x} : \frac{ln ( 2 x )}{x}

= \int \frac{ln ( 2 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} ( 2 x )}{2}

Add a constant to the solution

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

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