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

The integral from 0 to 3 of 1/(x+4) is ln(7)-2ln(2)

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integral from 0 to 3 of 1/(x+4)

\int_{0}^{3} \frac{1}{x + 4} d x

ln (7) - 2 ln (2)

Decimal

0.55961 \dots

Solution steps

\int_{0}^{3} \frac{1}{x + 4} d x

Apply u-substitution

= \int_{4}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{4}^{7}

Compute the boundaries:

ln (7) - 2 ln (2)

= ln (7) - 2 ln (2)

\int_{1}^{4} \frac{k}{x} d x

\int_{1}^{3} \frac{1}{x ^{5}} d x

\int_{0}^{4 - y^{2} - z^{2}} 1 d x

\int_{5}^{7} \frac{1}{x - 2} d x

\int_{0}^{1} x^{3} - 1 d x

What is the integral from 0 to 3 of 1/(x+4) ?
The integral from 0 to 3 of 1/(x+4) is ln(7)-2ln(2)