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

The integral of x/(3x-2) is 1/9 (3x-2+2ln|3x-2|)+C

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integral of x/(3x-2)

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Expand

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

= \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

= \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 ∣

= \frac{1}{9} (u + 2 ln ∣ u ∣)

Substitute back

u = 3 x - 2

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

Add a constant to the solution

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

f (x) = \frac{- x ^{2} - 1}{( x ^{2} - 1 ) ^{2}}

\int tan^{3} (19 π x) d x

\frac{1}{1 + x}

\int x - y d x

x \to \infty lim (x^{3} sin (\frac{1}{x}))

What is the integral of x/(3x-2) ?
The integral of x/(3x-2) is 1/9 (3x-2+2ln|3x-2|)+C