What is the integral of (2x-7)/(6x^2-5x+1) ?

The integral of (2x-7)/(6x^2-5x+1) is 19/3 ln|3x-1|-6ln|2x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x-7)/(6x^2-5x+1)

\int \frac{2 x - 7}{6 x ^{2} - 5 x + 1} d x

\frac{19}{3} ln ∣ 3 x - 1 ∣ - 6 ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{2 x - 7}{6 x ^{2} - 5 x + 1} d x

Take the partial fraction of

\frac{2 x - 7}{6 x ^{2} - 5 x + 1} : \frac{19}{3 x - 1} - \frac{12}{2 x - 1}

= \int \frac{19}{3 x - 1} - \frac{12}{2 x - 1} d x

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{19}{3 x - 1} d x - \int \frac{12}{2 x - 1} d x

\int \frac{19}{3 x - 1} d x = \frac{19}{3} ln ∣ 3 x - 1 ∣

\int \frac{12}{2 x - 1} d x = 6 ln ∣ 2 x - 1 ∣

= \frac{19}{3} ln ∣ 3 x - 1 ∣ - 6 ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{19}{3} ln ∣ 3 x - 1 ∣ - 6 ln ∣ 2 x - 1 ∣ + C

\int (6 x^{2} - 2 x) d x

\int 2 x cos (x^{2} - 5) d x

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

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

\int 4 x^{3} - 3 x^{2} + 2 x - 1 d x

What is the integral of (2x-7)/(6x^2-5x+1) ?
The integral of (2x-7)/(6x^2-5x+1) is 19/3 ln|3x-1|-6ln|2x-1|+C