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

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

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integral from 2 to 3 of 1/(x+1)

\int_{2}^{3} \frac{1}{x + 1} d x

2 ln (2) - ln (3)

Decimal

0.28768 \dots

Solution steps

\int_{2}^{3} \frac{1}{x + 1} d x

Apply u-substitution

= \int_{3}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{4}

Compute the boundaries:

2 ln (2) - ln (3)

= 2 ln (2) - ln (3)

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