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

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

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integral from-e-1 to-2 of 1/(x+1)

\int_{- e - 1}^{- 2} \frac{1}{x + 1} d x

- 1

Solution steps

\int_{- e - 1}^{- 2} \frac{1}{x + 1} d x

Apply u-substitution

: \int_{- e}^{- 1} \frac{1}{u} d u

= \int_{- e}^{- 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- e}^{- 1}

Compute the boundaries:

- 1

= - 1

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