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

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

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

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

2 ln (2)

Decimal

1.38629 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{4}

Compute the boundaries:

2 ln (2)

= 2 ln (2)

\int_{0}^{2} e^{x} d x

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\int_{- 1}^{2} k x^{2} d x

\int_{- 1}^{0} \frac{1}{x ^{4}} d x

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