What is the integral from 0 to x of 1/(1+x) ?

The integral from 0 to x of 1/(1+x) is ln|x+1|

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integral from 0 to x of 1/(1+x)

\int_{0}^{x} \frac{1}{1 + x} d x

ln ∣ x + 1 ∣

Solution steps

\int_{0}^{x} \frac{1}{1 + x} d x

Apply u-substitution

: \int_{1}^{1 + x} \frac{1}{u} d u

= \int_{1}^{1 + x} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{1 + x}

Compute the boundaries:

ln ∣ x + 1 ∣

= ln ∣ x + 1 ∣

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What is the integral from 0 to x of 1/(1+x) ?
The integral from 0 to x of 1/(1+x) is ln|x+1|