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

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

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

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

ln ∣ ln (4 x - 1) ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Apply u-substitution

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 4 \cdot \frac{1}{4} ln ∣ v ∣

Substitute back

= 4 \cdot \frac{1}{4} ln ∣ ln (4 x - 1) ∣

Simplify

4 \cdot \frac{1}{4} ln ∣ ln (4 x - 1) ∣ : ln ∣ ln (4 x - 1) ∣

= ln ∣ ln (4 x - 1) ∣

Add a constant to the solution

= ln ∣ ln (4 x - 1) ∣ + C

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