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

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

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

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

ln ∣ ln (2 x + 4) ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 u ln ( u )} d u

= 2 \cdot \int \frac{1}{2 u ln ( u )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u ln ( u )} d u

Apply u-substitution

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

= 2 \cdot \frac{1}{2} ln ∣ ln (2 x + 4) ∣

Simplify

2 \cdot \frac{1}{2} ln ∣ ln (2 x + 4) ∣ : ln ∣ ln (2 x + 4) ∣

= ln ∣ ln (2 x + 4) ∣

Add a constant to the solution

= ln ∣ ln (2 x + 4) ∣ + C

\int x^{2} y + tanh (x) d y

x \to 1 lim (1 - 5)

x \to \infty lim (\frac{12}{x})

\frac{d}{d t} (4 t^{2})

(2 x y^{2} + 5 y) + (2 x^{2} y + 5 x) y^{^{'}} = 0

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