What is the integral from 2 to 4 of (4x)/(4x-7) ?

The integral from 2 to 4 of (4x)/(4x-7) is 1/4 (8+14ln(3))

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integral from 2 to 4 of (4x)/(4x-7)

\int_{2}^{4} \frac{4 x}{4 x - 7} d x

\frac{1}{4} (8 + 14 ln (3))

Decimal Notation

5.84514 \dots

Solution steps

\int_{2}^{4} \frac{4 x}{4 x - 7} d x

Take the constant out:

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

= 4 \cdot \int_{2}^{4} \frac{x}{4 x - 7} d x

Apply u-substitution

: \int_{1}^{9} \frac{u + 7}{16 u} d u

= 4 \cdot \int_{1}^{9} \frac{u + 7}{16 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}{16} \cdot \int_{1}^{9} \frac{u + 7}{u} d u

Expand

\frac{u + 7}{u} : 1 + \frac{7}{u}

= 4 \cdot \frac{1}{16} \cdot \int_{1}^{9} 1 + \frac{7}{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}{16} (\int_{1}^{9} 1 d u + \int_{1}^{9} \frac{7}{u} d u)

\int_{1}^{9} 1 d u = 8

\int_{1}^{9} \frac{7}{u} d u = 14 ln (3)

= 4 \cdot \frac{1}{16} (8 + 14 ln (3))

Simplify

4 \cdot \frac{1}{16} (8 + 14 ln (3)) : \frac{1}{4} (8 + 14 ln (3))

= \frac{1}{4} (8 + 14 ln (3))

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What is the integral from 2 to 4 of (4x)/(4x-7) ?
The integral from 2 to 4 of (4x)/(4x-7) is 1/4 (8+14ln(3))