What is the integral of (ln(2x))/(ln(4x)) 1/x ?

The integral of (ln(2x))/(ln(4x)) 1/x is ln(4x)-ln(2)ln|ln(4x)|+C

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integral of (ln(2x))/(ln(4x)) 1/x

\int \frac{ln ( 2 x )}{ln ( 4 x )} \frac{1}{x} d x

ln (4 x) - ln (2) ln ∣ ln (4 x) ∣ + C

Solution steps

\int \frac{ln ( 2 x )}{ln ( 4 x )} \cdot \frac{1}{x} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{v - ln ( 2 )}{v} d v

= \int \frac{v - ln ( 2 )}{v} d v

Expand

\frac{v - ln ( 2 )}{v} : 1 - \frac{ln ( 2 )}{v}

= \int 1 - \frac{ln ( 2 )}{v} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d v - \int \frac{ln ( 2 )}{v} d v

\int 1 d v = v

\int \frac{ln ( 2 )}{v} d v = ln (2) ln ∣ v ∣

= v - ln (2) ln ∣ v ∣

Substitute back

= ln (4 x) - ln (2) ln ∣ ln (4 x) ∣

Add a constant to the solution

= ln (4 x) - ln (2) ln ∣ ln (4 x) ∣ + C

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What is the integral of (ln(2x))/(ln(4x)) 1/x ?
The integral of (ln(2x))/(ln(4x)) 1/x is ln(4x)-ln(2)ln|ln(4x)|+C