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

The integral of 3/((3x-1)ln(3x-1)) is ln|ln(3x-1)|+C

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integral of 3/((3x-1)ln(3x-1))

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

ln ∣ ln (3 x - 1) ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{3 u ln ( u )} d u

= 3 \cdot \int \frac{1}{3 u ln ( u )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int \frac{1}{u ln ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= 3 \cdot \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 3 \cdot \frac{1}{3} ln ∣ v ∣

Substitute back

= 3 \cdot \frac{1}{3} ln ∣ ln (3 x - 1) ∣

Simplify

3 \cdot \frac{1}{3} ln ∣ ln (3 x - 1) ∣ : ln ∣ ln (3 x - 1) ∣

= ln ∣ ln (3 x - 1) ∣

Add a constant to the solution

= ln ∣ ln (3 x - 1) ∣ + C

\int sin^{6} (a x) d x

\int cos^{2} (v) d v

\int (2 x + 15) e^{0.25 x} d x

x \to \infty lim (1^{0})

\frac{1}{4} x^{2}, 2 x^{2}, 3 - x

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