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

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

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

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

ln ∣ ln (x + 3) ∣ + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= ln ∣ v ∣

Substitute back

= ln ∣ ln (x + 3) ∣

Add a constant to the solution

= ln ∣ ln (x + 3) ∣ + C

e^{a x}

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

\int x ln (x + 2) d x

\int \frac{1}{t ^{2} - 9} d t

\frac{\partial}{\partial x} ((e^{- 6 x}) cos (y))

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