What is the integral from-6 to-3 of 1/(x+2) ?

The integral from-6 to-3 of 1/(x+2) is -2ln(2)

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integral from-6 to-3 of 1/(x+2)

\int_{- 6}^{- 3} \frac{1}{x + 2} d x

- 2 ln (2)

Decimal Notation

- 1.38629 \dots

Solution steps

\int_{- 6}^{- 3} \frac{1}{x + 2} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 4}^{- 1}

Compute the boundaries:

- 2 ln (2)

= - 2 ln (2)

\int_{0.75}^{0.9} 3 x^{2} d x

\int_{1}^{l n (2)} x e^{4 x} d x

\int_{0}^{6} 2 x 4 x + 1 d x

\int_{0}^{1} (- 1 + \frac{π x}{2}) d x

\int_{- 2}^{0} \frac{4}{( x + 2 ) ^{2}} d x

What is the integral from-6 to-3 of 1/(x+2) ?
The integral from-6 to-3 of 1/(x+2) is -2ln(2)