What is the integral of (4x)/(4x-2) ?

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

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integral of (4x)/(4x-2)

\int \frac{4 x}{4 x - 2} d x

\frac{1}{2} (2 x - 1 + ln (2 ∣ 2 x - 1 ∣)) + C

Solution steps

\int \frac{4 x}{4 x - 2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{x}{4 x - 2} d x

Apply u-substitution

: \int \frac{u + 2}{16 u} d u

= 4 \cdot \int \frac{u + 2}{16 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{16} \cdot \int \frac{u + 2}{u} d u

Expand

\frac{u + 2}{u} : 1 + \frac{2}{u}

= 4 \cdot \frac{1}{16} \cdot \int 1 + \frac{2}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{16} (\int 1 d u + \int \frac{2}{u} d u)

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= 4 \cdot \frac{1}{16} (u + 2 ln ∣ u ∣)

Substitute back

u = 4 x - 2

= 4 \cdot \frac{1}{16} (4 x - 2 + 2 ln ∣ 4 x - 2 ∣)

Simplify

4 \cdot \frac{1}{16} (4 x - 2 + 2 ln ∣ 4 x - 2 ∣) : \frac{1}{2} (2 x - 1 + ln (2 ∣ 2 x - 1 ∣))

= \frac{1}{2} (2 x - 1 + ln (2 ∣ 2 x - 1 ∣))

Add a constant to the solution

= \frac{1}{2} (2 x - 1 + ln (2 ∣ 2 x - 1 ∣)) + C

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