What is the integral from 0 to 4 of 1/(x+1) ?

The integral from 0 to 4 of 1/(x+1) is ln(5)

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integral from 0 to 4 of 1/(x+1)

\int_{0}^{4} \frac{1}{x + 1} d x

ln (5)

Solution steps

\int_{0}^{4} \frac{1}{x + 1} d x

Apply u-substitution

: \int_{1}^{5} \frac{1}{u} d u

= \int_{1}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

= ln (5)

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What is the integral from 0 to 4 of 1/(x+1) ?
The integral from 0 to 4 of 1/(x+1) is ln(5)