What is the integral of (2x)/(3x-2) ?

The integral of (2x)/(3x-2) is 2/9 (3x-2+2ln|3x-2|)+C

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integral of (2x)/(3x-2)

\int \frac{2 x}{3 x - 2} d x

\frac{2}{9} (3 x - 2 + 2 ln ∣ 3 x - 2 ∣) + C

Solution steps

\int \frac{2 x}{3 x - 2} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{3 x - 2} d x

Apply u-substitution

: \int \frac{u + 2}{9 u} d u

= 2 \cdot \int \frac{u + 2}{9 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \int \frac{u + 2}{u} d u

Expand

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

= 2 \cdot \frac{1}{9} \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

= 2 \cdot \frac{1}{9} (\int 1 d u + \int \frac{2}{u} d u)

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= 2 \cdot \frac{1}{9} (u + 2 ln ∣ u ∣)

Substitute back

u = 3 x - 2

= 2 \cdot \frac{1}{9} (3 x - 2 + 2 ln ∣ 3 x - 2 ∣)

Simplify

2 \cdot \frac{1}{9} (3 x - 2 + 2 ln ∣ 3 x - 2 ∣) : \frac{2}{9} (3 x - 2 + 2 ln ∣ 3 x - 2 ∣)

= \frac{2}{9} (3 x - 2 + 2 ln ∣ 3 x - 2 ∣)

Add a constant to the solution

= \frac{2}{9} (3 x - 2 + 2 ln ∣ 3 x - 2 ∣) + C

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What is the integral of (2x)/(3x-2) ?
The integral of (2x)/(3x-2) is 2/9 (3x-2+2ln|3x-2|)+C