What is the integral from 2 to 5 of 1/(3u+7) ?

The integral from 2 to 5 of 1/(3u+7) is 1/3 (ln(22)-ln(13))

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integral from 2 to 5 of 1/(3u+7)

\int_{2}^{5} \frac{1}{3 u + 7} d u

\frac{1}{3} (ln (22) - ln (13))

Decimal Notation

0.17536 \dots

Solution steps

\int_{2}^{5} \frac{1}{3 u + 7} d u

Apply u-substitution

: \int_{13}^{22} \frac{1}{3 v} d v

= \int_{13}^{22} \frac{1}{3 v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{13}^{22} \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} [ln ∣ v ∣]_{13}^{22}

Compute the boundaries:

ln (22) - ln (13)

= \frac{1}{3} (ln (22) - ln (13))

\int_{2.9}^{4} \frac{1}{3} d x

\int_{0}^{2} (x e^{x}) d x

\int_{2}^{e} \frac{1}{x ( ln ( x ) ) ^{2}} d x

\int_{- 1}^{1} sec^{2} (x) d x

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

What is the integral from 2 to 5 of 1/(3u+7) ?
The integral from 2 to 5 of 1/(3u+7) is 1/3 (ln(22)-ln(13))