What is the integral from 0 to 10 of (24x)/(2x+3) ?

The integral from 0 to 10 of (24x)/(2x+3) is 6(20-3(ln(23)-ln(3)))

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integral from 0 to 10 of (24x)/(2x+3)

\int_{0}^{10} \frac{24 x}{2 x + 3} d x

6 (20 - 3 (ln (23) - ln (3)))

Decimal

83.33612 \dots

Solution steps

\int_{0}^{10} \frac{24 x}{2 x + 3} d x

Take the constant out:

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

= 24 \cdot \int_{0}^{10} \frac{x}{2 x + 3} d x

Apply u-substitution

: \int_{3}^{23} \frac{u - 3}{4 u} d u

= 24 \cdot \int_{3}^{23} \frac{u - 3}{4 u} d u

Take the constant out:

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

= 24 \cdot \frac{1}{4} \cdot \int_{3}^{23} \frac{u - 3}{u} d u

Expand

\frac{u - 3}{u} : 1 - \frac{3}{u}

= 24 \cdot \frac{1}{4} \cdot \int_{3}^{23} 1 - \frac{3}{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

= 24 \cdot \frac{1}{4} (\int_{3}^{23} 1 d u - \int_{3}^{23} \frac{3}{u} d u)

\int_{3}^{23} 1 d u = 20

\int_{3}^{23} \frac{3}{u} d u = 3 (ln (23) - ln (3))

= 24 \cdot \frac{1}{4} (20 - 3 (ln (23) - ln (3)))

Simplify

24 \cdot \frac{1}{4} (20 - 3 (ln (23) - ln (3))) : 6 (20 - 3 (ln (23) - ln (3)))

= 6 (20 - 3 (ln (23) - ln (3)))

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What is the integral from 0 to 10 of (24x)/(2x+3) ?
The integral from 0 to 10 of (24x)/(2x+3) is 6(20-3(ln(23)-ln(3)))