What is the integral from 0 to 1 of (1/(x+2)) ?

The integral from 0 to 1 of (1/(x+2)) is ln(3)-ln(2)

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integral from 0 to 1 of (1/(x+2))

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

ln (3) - ln (2)

Decimal

0.40546 \dots

Solution steps

\int_{0}^{1} \frac{1}{x + 2} d x

Apply u-substitution

: \int_{2}^{3} \frac{1}{u} d u

= \int_{2}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{3}

Compute the boundaries:

ln (3) - ln (2)

= ln (3) - ln (2)

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

\int_{0}^{\infty} x e^{- θ x} d x

\int_{1}^{2} (1 - x^{2}) d x

\int_{0}^{π} sin^{4} (t) d t

\int_{- 1}^{0} 3 t^{2} e^{(- t^{3} + 1)} d t

What is the integral from 0 to 1 of (1/(x+2)) ?
The integral from 0 to 1 of (1/(x+2)) is ln(3)-ln(2)