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

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

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

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

ln (2)

Decimal

0.69314 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Apply u-substitution

= - \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))

Simplify

= ln (2)

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