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

The integral of 1/(x^3-7x+6) is -1/4 ln|x-1|+1/5 ln|x-2|+1/20 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 1/(x^3-7x+6)

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

- \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{20} ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{3} - 7 x + 6} : - \frac{1}{4 ( x - 1 )} + \frac{1}{5 ( x - 2 )} + \frac{1}{20 ( x + 3 )}

= \int - \frac{1}{4 ( x - 1 )} + \frac{1}{5 ( x - 2 )} + \frac{1}{20 ( 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{1}{4 ( x - 1 )} d x + \int \frac{1}{5 ( x - 2 )} d x + \int \frac{1}{20 ( x + 3 )} d x

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

\int \frac{1}{5 ( x - 2 )} d x = \frac{1}{5} ln ∣ x - 2 ∣

\int \frac{1}{20 ( x + 3 )} d x = \frac{1}{20} ln ∣ x + 3 ∣

= - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{20} ln ∣ x + 3 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{20} ln ∣ x + 3 ∣ + C

\int \frac{4 x ^{2} - 4 x + 24}{x ^{3} + 3 x} d x

\int \frac{x e ^{x}}{9 + e ^{x}} d x

\int + \frac{4 y ^{2} - 7 y - 12}{y ( y + 2 ) ( y - 3 )} d y

\int \frac{1}{x ^{0.8}} d x

\int 4 x^{2} e^{3 x^{2}} d x

What is the integral of 1/(x^3-7x+6) ?
The integral of 1/(x^3-7x+6) is -1/4 ln|x-1|+1/5 ln|x-2|+1/20 ln|x+3|+C