What is the integral from 2 to 3 of 1/(3x-2) ?

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

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integral from 2 to 3 of 1/(3x-2)

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

\frac{1}{3} (ln (7) - 2 ln (2))

Decimal

0.18653 \dots

Solution steps

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

Apply u-substitution

= \int_{4}^{7} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{4}^{7}

Compute the boundaries:

ln (7) - 2 ln (2)

= \frac{1}{3} (ln (7) - 2 ln (2))

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