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

The integral from 1 to 3 of 1/(x+2) is ln(5)-ln(3)

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

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

ln (5) - ln (3)

Decimal

0.51082 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{5} \frac{1}{u} d u

= \int_{3}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{5}

Compute the boundaries:

ln (5) - ln (3)

= ln (5) - ln (3)

\int_{0}^{t} 0 d x

\int_{- \infty}^{2} \frac{1}{5 - 2 x} d x

\int_{0}^{1} (t - \frac{1}{2})^{2} d t

\int_{0}^{2} (4 + x^{2})^{\frac{3}{2}} d x

\int_{0}^{3} (2 x^{2}) d x

What is the integral from 1 to 3 of 1/(x+2) ?
The integral from 1 to 3 of 1/(x+2) is ln(5)-ln(3)