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

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

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

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

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

Decimal Notation

2.74653 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{1}{2 u} d u

= 5 \cdot \int_{1}^{3} \frac{1}{2 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}{2} \cdot \int_{1}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{3}

Simplify

5 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{3} : \frac{5}{2} [ln ∣ u ∣]_{1}^{3}

= \frac{5}{2} [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

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

\int 9 cos (3 x) d x

\int x (x + 2)^{\frac{1}{2}} d x

\frac{d}{d x} (3 x^{2} - 7)

(2 x y + 3 y^{2}) d x = (2 x y + x^{2}) d y

x \to 2 lim (\frac{x ^{3} - 8}{2 - x})

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