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

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

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

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

ln (7) - ln (3)

Solution steps

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

Apply u-substitution

: \int_{3}^{7} \frac{1}{u} d u

= \int_{3}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{7}

Compute the boundaries:

ln (7) - ln (3)

= ln (7) - ln (3)

\int_{0}^{1} - 6 x^{3} (3 x^{4} - 2)^{5} d x

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

\int_{0}^{5} x^{2} - 9 d x

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

\int_{0}^{\frac{π}{4}} (9 sec^{2} (x)) d x

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