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)

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_{1}^{3} (x + 3)^{3} d x

\int_{- 3}^{0} x + 3 + 9 - x^{2} d x

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

\int_{0}^{\frac{π}{2}} (cos (2 x)) d x

\int_{1}^{2} t 9 + 16 t^{2} 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)