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

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

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integral of 1/(2x-1)

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

\frac{1}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Apply u-substitution

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

= \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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 1

= \frac{1}{2} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x - 1 ∣ + C

\frac{d}{d x} (\frac{5 1 - x ^{3}}{6})

\int 13 arctan (x) d x

x \to + 0 lim (\frac{e ^{2 x} - 1}{sin ( 3 x )})

x \to 3 lim (x^{3} - 2 x^{2} + 1)

n = 1 \sum \infty \frac{30}{3 ^{n}}