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

The integral of 1/(2(2x-1)) is 1/4 ln|2x-1|+C

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integral of 1/(2(2x-1))

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

\frac{1}{4} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

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}{2 x - 1} d x

Apply u-substitution

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

= \frac{1}{2} \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

= \frac{1}{2} \cdot \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} \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 1

= \frac{1}{2} \cdot \frac{1}{2} ln ∣ 2 x - 1 ∣

Simplify

\frac{1}{2} \cdot \frac{1}{2} ln ∣ 2 x - 1 ∣ : \frac{1}{4} ln ∣ 2 x - 1 ∣

= \frac{1}{4} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 2 x - 1 ∣ + C

\int + cos (x) cos (7 x) d x

\int \frac{w - 1}{2 w ^{2} - 2 w + 1} d w

e^{(2 x)}

\int \frac{sin ( h ) 7 x}{1 + cos ( h ^{2} ) 7 x} d x

What is the integral of 1/(2(2x-1)) ?
The integral of 1/(2(2x-1)) is 1/4 ln|2x-1|+C