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

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

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

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

ln (2)

Solution steps

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

Apply u-substitution

: \int_{2}^{4} \frac{1}{u} d u

= \int_{2}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{4}

Compute the boundaries:

ln (2)

= ln (2)

x \to 0 lim (\frac{( cos ( 11 x ) - cos ( 6 x ) )}{x})

f (x) = x^{3} sin (x)

\frac{d}{d x} (\frac{x ^{2} - 6 x + 1}{( x ^{2} - 1 ) ^{2}})

\int cos (x) (5 + 5 sin^{2} (x)) d x

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