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

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

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

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

Ei (ln (x + 1)) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{e ^{v}}{v} d v

= \int \frac{e ^{v}}{v} d v

Use the nonelementary integral:

\int \frac{e ^{x}}{x} d x = Ei (x)

= Ei (v)

Substitute back

= Ei (ln (x + 1))

Add a constant to the solution

= Ei (ln (x + 1)) + C

\int \frac{1}{x ^{2} 10 x + 1} d x

\int e^{θ} sin (θ) d θ

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

n = 1 \sum \infty \frac{( ( - 3 ) ^{n - 1} )}{5 ^{n}}

f (x) = 4 (x - 4)^{2} (x - 1)

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