What is the integral from 3 to 5 of (4x)/(3x-2) ?

The integral from 3 to 5 of (4x)/(3x-2) is 4/9 (6+2(ln(13)-ln(7)))

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integral from 3 to 5 of (4x)/(3x-2)

\int_{3}^{5} \frac{4 x}{3 x - 2} d x

\frac{4}{9} (6 + 2 (ln (13) - ln (7)))

Decimal

3.21692 \dots

Solution steps

\int_{3}^{5} \frac{4 x}{3 x - 2} d x

Take the constant out:

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

= 4 \cdot \int_{3}^{5} \frac{x}{3 x - 2} d x

Apply u-substitution

: \int_{7}^{13} \frac{u + 2}{9 u} d u

= 4 \cdot \int_{7}^{13} \frac{u + 2}{9 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{9} \cdot \int_{7}^{13} \frac{u + 2}{u} d u

Expand

\frac{u + 2}{u} : 1 + \frac{2}{u}

= 4 \cdot \frac{1}{9} \cdot \int_{7}^{13} 1 + \frac{2}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{9} (\int_{7}^{13} 1 d u + \int_{7}^{13} \frac{2}{u} d u)

\int_{7}^{13} 1 d u = 6

\int_{7}^{13} \frac{2}{u} d u = 2 (ln (13) - ln (7))

= 4 \cdot \frac{1}{9} (6 + 2 (ln (13) - ln (7)))

Simplify

4 \cdot \frac{1}{9} (6 + 2 (ln (13) - ln (7))) : \frac{4}{9} (6 + 2 (ln (13) - ln (7)))

= \frac{4}{9} (6 + 2 (ln (13) - ln (7)))

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What is the integral from 3 to 5 of (4x)/(3x-2) ?
The integral from 3 to 5 of (4x)/(3x-2) is 4/9 (6+2(ln(13)-ln(7)))