What is the integral of (6x^2-11x+5)/(3x-1) ?

The integral of (6x^2-11x+5)/(3x-1) is x^2-3x+2/3 ln|3x-1|-1/3+11/9+C

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integral of (6x^2-11x+5)/(3x-1)

\int \frac{6 x ^{2} - 11 x + 5}{3 x - 1} d x

x^{2} - 3 x + \frac{2}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{3} + \frac{11}{9} + C

Solution steps

\int \frac{6 x ^{2} - 11 x + 5}{3 x - 1} d x

Expand

\frac{6 x ^{2} - 11 x + 5}{3 x - 1} : \frac{6 x ^{2}}{3 x - 1} - \frac{11 x}{3 x - 1} + \frac{5}{3 x - 1}

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{6 x ^{2}}{3 x - 1} d x - \int \frac{11 x}{3 x - 1} d x + \int \frac{5}{3 x - 1} d x

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

\int \frac{11 x}{3 x - 1} d x = \frac{11}{9} (3 x - 1 + ln ∣ 3 x - 1 ∣)

\int \frac{5}{3 x - 1} d x = \frac{5}{3} ln ∣ 3 x - 1 ∣

= \frac{1}{9} (9 x^{2} + 6 x + 2 ln ∣ 3 x - 1 ∣ - 3) - \frac{11}{9} (3 x - 1 + ln ∣ 3 x - 1 ∣) + \frac{5}{3} ln ∣ 3 x - 1 ∣

Simplify

\frac{1}{9} (9 x^{2} + 6 x + 2 ln ∣ 3 x - 1 ∣ - 3) - \frac{11}{9} (3 x - 1 + ln ∣ 3 x - 1 ∣) + \frac{5}{3} ln ∣ 3 x - 1 ∣ : x^{2} - 3 x + \frac{2}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{3} + \frac{11}{9}

= x^{2} - 3 x + \frac{2}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{3} + \frac{11}{9}

Add a constant to the solution

= x^{2} - 3 x + \frac{2}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{3} + \frac{11}{9} + C

\int_{- \infty}^{2} \frac{2}{x ^{2} + 4} d x

f (x) = 2 x - x^{2}

\int cos (x) sin^{5} (x) d x

3 - x, - 4 \leq x \leq 2

\frac{d}{d x} ((1 - x^{2})^{\frac{1}{3}})

What is the integral of (6x^2-11x+5)/(3x-1) ?
The integral of (6x^2-11x+5)/(3x-1) is x^2-3x+2/3 ln|3x-1|-1/3+11/9+C