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

The integral from 0 to 2 of 1/(x+3) is ln(5)-ln(3)

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integral from 0 to 2 of 1/(x+3)

\int_{0}^{2} \frac{1}{x + 3} d x

ln (5) - ln (3)

Decimal Notation

0.51082 \dots

Solution steps

\int_{0}^{2} \frac{1}{x + 3} 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)

cos (2 y)

\int \frac{1}{y ^{2} - 1} d y

n = 2 \sum \infty \frac{1}{n ( n + 2 )}

x y^{^{'}} + 2 y = 9 x

n = 1 \sum \infty \frac{( 4 ) ^{n - 1}}{5 ^{n}}

What is the integral from 0 to 2 of 1/(x+3) ?
The integral from 0 to 2 of 1/(x+3) is ln(5)-ln(3)