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

The integral from 1 to 2 of (5x)/(2x-1) is 5/4 (2+ln(3))

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

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

\frac{5}{4} (2 + ln (3))

Decimal Notation

3.87326 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{3} \frac{u + 1}{4 u} d u

= 5 \cdot \int_{1}^{3} \frac{u + 1}{4 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int_{1}^{3} \frac{u + 1}{u} d u

Expand

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

= 5 \cdot \frac{1}{4} \cdot \int_{1}^{3} 1 + \frac{1}{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

= 5 \cdot \frac{1}{4} (\int_{1}^{3} 1 d u + \int_{1}^{3} \frac{1}{u} d u)

\int_{1}^{3} 1 d u = 2

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

= 5 \cdot \frac{1}{4} (2 + ln (3))

Simplify

5 \cdot \frac{1}{4} (2 + ln (3)) : \frac{5}{4} (2 + ln (3))

= \frac{5}{4} (2 + ln (3))

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What is the integral from 1 to 2 of (5x)/(2x-1) ?
The integral from 1 to 2 of (5x)/(2x-1) is 5/4 (2+ln(3))