What is the integral from 0 to 12 of 1/(x+12) ?

The integral from 0 to 12 of 1/(x+12) is ln(2)

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integral from 0 to 12 of 1/(x+12)

\int_{0}^{12} \frac{1}{x + 12} d x

ln (2)

Solution steps

\int_{0}^{12} \frac{1}{x + 12} d x

Apply u-substitution

: \int_{12}^{24} \frac{1}{u} d u

= \int_{12}^{24} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{12}^{24}

Compute the boundaries:

ln (24) - ln (12)

= ln (24) - ln (12)

Simplify

= ln (2)

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What is the integral from 0 to 12 of 1/(x+12) ?
The integral from 0 to 12 of 1/(x+12) is ln(2)