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

The integral from 0 to 2 of 1/(x+1) is ln(3)

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

\int_{0}^{2} \frac{1}{x + 1} d x

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

\int_{0}^{2} \frac{1}{x + 1} d x

Apply u-substitution

: \int_{1}^{3} \frac{1}{u} d u

= \int_{1}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= ln (3)

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What is the integral from 0 to 2 of 1/(x+1) ?
The integral from 0 to 2 of 1/(x+1) is ln(3)