What is the integral from 1 to 7 of 1/(x+1) ?

The integral from 1 to 7 of 1/(x+1) is 2ln(2)

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integral from 1 to 7 of 1/(x+1)

\int_{1}^{7} \frac{1}{x + 1} d x

2 ln (2)

Solution steps

\int_{1}^{7} \frac{1}{x + 1} d x

Apply u-substitution

: \int_{2}^{8} \frac{1}{u} d u

= \int_{2}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{8}

Compute the boundaries:

2 ln (2)

= 2 ln (2)

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What is the integral from 1 to 7 of 1/(x+1) ?
The integral from 1 to 7 of 1/(x+1) is 2ln(2)