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

The integral of 5/(2x-1) is 5/2 ln|2x-1|+C

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integral of 5/(2x-1)

\int \frac{5}{2 x - 1} d x

\frac{5}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \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 \frac{1}{2 x - 1} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 5 \cdot \int \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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 1

= 5 \cdot \frac{1}{2} ln ∣ 2 x - 1 ∣

Simplify

5 \cdot \frac{1}{2} ln ∣ 2 x - 1 ∣ : \frac{5}{2} ln ∣ 2 x - 1 ∣

= \frac{5}{2} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{5}{2} ln ∣ 2 x - 1 ∣ + C

\int_{5}^{8} (x - 7)^{2} d x

\frac{d y}{d x} = \frac{1}{y ^{3}}

x \to - 2 + lim (\frac{x + 1}{x + 2})

\int \frac{1}{x 4 x ^{2} - 16} d x

y = 7 + 5 x^{2} - 2 x^{3}, (2, 11)