What is the integral of (1+x)^{-1} ?

The integral of (1+x)^{-1} is ln|1+x|+C

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integral of (1+x)^{-1}

\int (1 + x)^{- 1} d x

ln ∣ 1 + x ∣ + C

Solution steps

\int (1 + x)^{- 1} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 1 + x

= ln ∣ 1 + x ∣

Add a constant to the solution

= ln ∣ 1 + x ∣ + C

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