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

The integral from 0 to 1 of 1/(x+1) is ln(2)

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

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

ln (2)

Decimal

0.69314 \dots

Solution steps

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

Apply u-substitution

= \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

What is the integral from 0 to 1 of 1/(x+1) ?
The integral from 0 to 1 of 1/(x+1) is ln(2)