What is the integral from 0 to 2 of 1/(x+4) ?

The integral from 0 to 2 of 1/(x+4) is ln(3/2)

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integral from 0 to 2 of 1/(x+4)

\int_{0}^{2} \frac{1}{x + 4} d x

ln (\frac{3}{2})

Solution steps

\int_{0}^{2} \frac{1}{x + 4} d x

Apply u-substitution

: \int_{4}^{6} \frac{1}{u} d u

= \int_{4}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{4}^{6}

Compute the boundaries:

ln (6) - 2 ln (2)

= ln (6) - 2 ln (2)

Simplify

= ln (\frac{3}{2})

\int_{1}^{6} t^{2} - 2 t - 8 d t

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

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

\int_{x^{2}}^{x} (x^{2} + y^{2}) d y

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

What is the integral from 0 to 2 of 1/(x+4) ?
The integral from 0 to 2 of 1/(x+4) is ln(3/2)