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

t^{2} \frac{d y}{d t} - t = 1 + y + y t

y^{^{''}} = ey

\frac{d}{d x} (e^{x + 5})

y = ln (x - 1), y = x - 2, x = 3

\frac{100}{s ( s ^{2} + 2 s + 6 )}

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