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

The integral of (ln(2x))/(ln(4x)) is (xln(2x))/(ln(4x))+(ln(2)x)/(ln(4x))-1/4 ln(2)\Ei(ln(4x))+C

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

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

\frac{x ln ( 2 x )}{ln ( 4 x )} + \frac{ln ( 2 ) x}{ln ( 4 x )} - \frac{1}{4} ln (2) Ei (ln (4 x)) + C

Solution steps

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

Apply u-substitution

: \int \frac{ln ( \frac{u}{2} )}{4 ln ( u )} d u

= \int \frac{ln ( \frac{u}{2} )}{4 ln ( u )} d u

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

Apply Integration By Parts

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

= \frac{1}{4} (\frac{u ln ( \frac{u}{2} )}{ln ( u )} - \int \frac{ln ( 2 )}{ln ^{2} ( u )} d u)

\int \frac{ln ( 2 )}{ln ^{2} ( u )} d u = ln (2) (- \frac{u}{ln ( u )} + Ei (ln (u)))

= \frac{1}{4} (\frac{u ln ( \frac{u}{2} )}{ln ( u )} - ln (2) (- \frac{u}{ln ( u )} + Ei (ln (u))))

Substitute back

u = 4 x

= \frac{1}{4} (\frac{4 x ln ( \frac{4 x}{2} )}{ln ( 4 x )} - ln (2) (- \frac{4 x}{ln ( 4 x )} + Ei (ln (4 x))))

Simplify

\frac{1}{4} (\frac{4 x ln ( \frac{4 x}{2} )}{ln ( 4 x )} - ln (2) (- \frac{4 x}{ln ( 4 x )} + Ei (ln (4 x)))) : \frac{x ln ( 2 x )}{ln ( 4 x )} + \frac{ln ( 2 ) x}{ln ( 4 x )} - \frac{1}{4} ln (2) Ei (ln (4 x))

= \frac{x ln ( 2 x )}{ln ( 4 x )} + \frac{ln ( 2 ) x}{ln ( 4 x )} - \frac{1}{4} ln (2) Ei (ln (4 x))

Add a constant to the solution

= \frac{x ln ( 2 x )}{ln ( 4 x )} + \frac{ln ( 2 ) x}{ln ( 4 x )} - \frac{1}{4} ln (2) Ei (ln (4 x)) + C

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