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

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

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

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

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{1}{u} d u

= \int_{1}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= ln (3)

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

\frac{d ^{4} y}{d x ^{4}} - 23 \frac{d ^{2} y}{d x ^{2}} - 50 y = 0

5 x + \frac{4}{x}

x \to 0 lim (- (\frac{1}{x} - \frac{1}{∣ x ∣}))

x \to 0 lim (sin (2 x))

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