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

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

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

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

ln ∣ ln (x + 1) ∣ + C

Solution steps

\int \frac{1}{( x + 1 ) ln ( x + 1 )} 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 + 1) ∣

Add a constant to the solution

= ln ∣ ln (x + 1) ∣ + C

y = 3 - 2 x + 4 x^{3}, at x = - 1

\frac{d}{d x} (ln (x^{16}))

y^{^{'}} = x y + x + y + 1

\int_{0}^{1} 7 (x^{2} + 1) e^{- x} d x

\int \frac{x ^{3} - 8 x}{x ^{2}} d x

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