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

The integral of 1/(x^3+2x^2-5x-6) is -1/6 ln|x+1|+1/15 ln|x-2|+1/10 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+2x^2-5x-6)

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

- \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{15} ln ∣ x - 2 ∣ + \frac{1}{10} ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{3} + 2 x ^{2} - 5 x - 6} : - \frac{1}{6 ( x + 1 )} + \frac{1}{15 ( x - 2 )} + \frac{1}{10 ( x + 3 )}

= \int - \frac{1}{6 ( x + 1 )} + \frac{1}{15 ( x - 2 )} + \frac{1}{10 ( 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}{6 ( x + 1 )} d x + \int \frac{1}{15 ( x - 2 )} d x + \int \frac{1}{10 ( x + 3 )} d x

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

\int \frac{1}{15 ( x - 2 )} d x = \frac{1}{15} ln ∣ x - 2 ∣

\int \frac{1}{10 ( x + 3 )} d x = \frac{1}{10} ln ∣ x + 3 ∣

= - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{15} ln ∣ x - 2 ∣ + \frac{1}{10} ln ∣ x + 3 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{15} ln ∣ x - 2 ∣ + \frac{1}{10} ln ∣ x + 3 ∣ + C

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

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

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

\int \frac{1}{25 + 4 x ^{2}} d x

\int \frac{8 x ^{3} - 7 x + 5}{x ^{3} - x ^{2}} d x

What is the integral of 1/(x^3+2x^2-5x-6) ?
The integral of 1/(x^3+2x^2-5x-6) is -1/6 ln|x+1|+1/15 ln|x-2|+1/10 ln|x+3|+C