What is the integral of (6x^2-11x+7)/(3x-1) ?

The integral of (6x^2-11x+7)/(3x-1) is x^2-3x+4/3 ln|3x-1|-1/3+11/9+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x^2-11x+7)/(3x-1)

\int \frac{6 x ^{2} - 11 x + 7}{3 x - 1} d x

x^{2} - 3 x + \frac{4}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{3} + \frac{11}{9} + C

Solution steps

\int \frac{6 x ^{2} - 11 x + 7}{3 x - 1} d x

Expand

\frac{6 x ^{2} - 11 x + 7}{3 x - 1} : \frac{6 x ^{2}}{3 x - 1} - \frac{11 x}{3 x - 1} + \frac{7}{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{7}{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{7}{3 x - 1} d x = \frac{7}{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{7}{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{7}{3} ln ∣ 3 x - 1 ∣ : x^{2} - 3 x + \frac{4}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{3} + \frac{11}{9}

= x^{2} - 3 x + \frac{4}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{3} + \frac{11}{9}

Add a constant to the solution

= x^{2} - 3 x + \frac{4}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{3} + \frac{11}{9} + C

\int \frac{1 x - 2}{x + 4} d x

\int e^{- s i n (x)} sin (x) cos (x) d x

\int 2 x tan^{2} (x^{2}) d x

\int (e^{x} - x)^{2} d x

\int \frac{1}{( x ^{2} + a ) ^{2}} d x

What is the integral of (6x^2-11x+7)/(3x-1) ?
The integral of (6x^2-11x+7)/(3x-1) is x^2-3x+4/3 ln|3x-1|-1/3+11/9+C