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

The integral from-1 to 7 of 1/(x+1) is diverges

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{8} \frac{1}{u} d u

= \int_{0}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{8}

Compute the boundaries:

diverges

= diverges

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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 diverges