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

The integral from 1 to 3 of 3/(7x-2) is 3/7 (ln(19)-ln(5))

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

\int_{1}^{3} \frac{3}{7 x - 2} d x

\frac{3}{7} (ln (19) - ln (5))

Decimal

0.57214 \dots

Solution steps

\int_{1}^{3} \frac{3}{7 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}^{3} \frac{1}{7 x - 2} d x

Apply u-substitution

: \int_{5}^{19} \frac{1}{7 u} d u

= 3 \cdot \int_{5}^{19} \frac{1}{7 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}{7} \cdot \int_{5}^{19} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{7} [ln ∣ u ∣]_{5}^{19}

Simplify

3 \cdot \frac{1}{7} [ln ∣ u ∣]_{5}^{19} : \frac{3}{7} [ln ∣ u ∣]_{5}^{19}

= \frac{3}{7} [ln ∣ u ∣]_{5}^{19}

Compute the boundaries:

ln (19) - ln (5)

= \frac{3}{7} (ln (19) - ln (5))

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What is the integral from 1 to 3 of 3/(7x-2) ?
The integral from 1 to 3 of 3/(7x-2) is 3/7 (ln(19)-ln(5))