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

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

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

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

ln (3) - ln (2)

Decimal Notation

0.40546 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{3} \frac{1}{u} d u

= \int_{2}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{3}

Compute the boundaries:

ln (3) - ln (2)

= ln (3) - ln (2)

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\int_{- 4}^{4} (4 - ∣ x ∣) d x

\int_{0}^{e - 1} ln (1 + x) d x

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