What is the integral from-10 to-6 of 1/(x+2) ?

The integral from-10 to-6 of 1/(x+2) is -ln(2)

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integral from-10 to-6 of 1/(x+2)

\int_{- 10}^{- 6} \frac{1}{x + 2} d x

- ln (2)

Decimal

- 0.69314 \dots

Solution steps

\int_{- 10}^{- 6} \frac{1}{x + 2} d x

Apply u-substitution

: \int_{- 8}^{- 4} \frac{1}{u} d u

= \int_{- 8}^{- 4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 8}^{- 4}

Compute the boundaries:

- ln (2)

= - ln (2)

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What is the integral from-10 to-6 of 1/(x+2) ?
The integral from-10 to-6 of 1/(x+2) is -ln(2)