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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the partial fraction of

\frac{2 x + 1}{x ^{3} - 7 x + 6} : - \frac{3}{4 ( x - 1 )} + \frac{1}{x - 2} - \frac{1}{4 ( x + 3 )}

= \int - \frac{3}{4 ( x - 1 )} + \frac{1}{x - 2} - \frac{1}{4 ( x + 3 )} 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{3}{4 ( x - 1 )} d x + \int \frac{1}{x - 2} d x - \int \frac{1}{4 ( x + 3 )} d x

\int \frac{3}{4 ( x - 1 )} d x = \frac{3}{4} ln ∣ x - 1 ∣

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

\int \frac{1}{4 ( x + 3 )} d x = \frac{1}{4} ln ∣ x + 3 ∣

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

Add a constant to the solution

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

\int 2 + 3^{x} d x

\int \frac{1}{x ^{5} + x ^{3}} d x

\int 8 sin (2 x + 90) d x

\int x a x^{2} + b d x

\int \frac{6}{( x - 7 ) ( x - 1 )} d x

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