What is the integral from 1 to 5 of 3/(3x-2) ?

The integral from 1 to 5 of 3/(3x-2) is ln(13)

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integral from 1 to 5 of 3/(3x-2)

\int_{1}^{5} \frac{3}{3 x - 2} d x

ln (13)

Solution steps

\int_{1}^{5} \frac{3}{3 x - 2} d x

Take the constant out:

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

= 3 \cdot \int_{1}^{5} \frac{1}{3 x - 2} d x

Apply u-substitution

: \int_{1}^{13} \frac{1}{3 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{13}

Simplify

3 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{13} : [ln ∣ u ∣]_{1}^{13}

= [ln ∣ u ∣]_{1}^{13}

Compute the boundaries:

ln (13)

= ln (13)

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What is the integral from 1 to 5 of 3/(3x-2) ?
The integral from 1 to 5 of 3/(3x-2) is ln(13)