What is the integral from 2 to 8 of 1/(2x+1) ?

The integral from 2 to 8 of 1/(2x+1) is 1/2 (ln(17)-ln(5))

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integral from 2 to 8 of 1/(2x+1)

\int_{2}^{8} \frac{1}{2 x + 1} d x

\frac{1}{2} (ln (17) - ln (5))

Decimal

0.61188 \dots

Solution steps

\int_{2}^{8} \frac{1}{2 x + 1} d x

Apply u-substitution

: \int_{5}^{17} \frac{1}{2 u} d u

= \int_{5}^{17} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{5}^{17} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{5}^{17}

Compute the boundaries:

ln (17) - ln (5)

= \frac{1}{2} (ln (17) - ln (5))

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What is the integral from 2 to 8 of 1/(2x+1) ?
The integral from 2 to 8 of 1/(2x+1) is 1/2 (ln(17)-ln(5))