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

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

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

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

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

Solution steps

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

Expand

\frac{3 x ^{2} - x}{3 x + 2} : \frac{3 x ^{2}}{3 x + 2} - \frac{x}{3 x + 2}

Apply the Sum Rule:

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

= \int \frac{3 x ^{2}}{3 x + 2} d x - \int \frac{x}{3 x + 2} d x

\int \frac{3 x ^{2}}{3 x + 2} d x = 3 (\frac{x ^{2}}{6} - \frac{2}{9} x + \frac{4}{27} ln ∣ 3 x + 2 ∣)

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

= 3 (\frac{x ^{2}}{6} - \frac{2}{9} x + \frac{4}{27} ln ∣ 3 x + 2 ∣) - \frac{1}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣)

Simplify

3 (\frac{x ^{2}}{6} - \frac{2}{9} x + \frac{4}{27} ln ∣ 3 x + 2 ∣) - \frac{1}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣) : \frac{x ^{2}}{2} - x + \frac{2}{3} ln ∣ 3 x + 2 ∣ - \frac{2}{9}

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

Add a constant to the solution

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

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