What is the integral of 3/((3x+4)ln(3x+4)) ?

The integral of 3/((3x+4)ln(3x+4)) is ln|ln(3x+4)|+C

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integral of 3/((3x+4)ln(3x+4))

\int \frac{3}{( 3 x + 4 ) ln ( 3 x + 4 )} d x

ln ∣ ln (3 x + 4) ∣ + C

Solution steps

\int \frac{3}{( 3 x + 4 ) ln ( 3 x + 4 )} 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 + 4 ) ln ( 3 x + 4 )} 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 + 4) ∣

Simplify

3 \cdot \frac{1}{3} ln ∣ ln (3 x + 4) ∣ : ln ∣ ln (3 x + 4) ∣

= ln ∣ ln (3 x + 4) ∣

Add a constant to the solution

= ln ∣ ln (3 x + 4) ∣ + C

x \to 0 lim (sin (3 x) cot (x))

x \to - 7 + lim (\frac{- x ^{2}}{x + 7})

x \to 1 lim (\frac{- 2 x ^{2}}{x - 1})

\int_{0}^{1} 2 ln (x) d x

\int lo g_{e} (x^{2} + 1 - x) d x

What is the integral of 3/((3x+4)ln(3x+4)) ?
The integral of 3/((3x+4)ln(3x+4)) is ln|ln(3x+4)|+C