What is the integral of 1/(x^2+9x+18) ?

The integral of 1/(x^2+9x+18) is 1/3 ln|x+3|-1/3 ln|x+6|+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^2+9x+18)

\int \frac{1}{x ^{2} + 9 x + 18} d x

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

Solution steps

\int \frac{1}{x ^{2} + 9 x + 18} d x

Take the partial fraction of

\frac{1}{x ^{2} + 9 x + 18} : \frac{1}{3 ( x + 3 )} - \frac{1}{3 ( x + 6 )}

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

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

\int \frac{1}{3 ( x + 6 )} d x = \frac{1}{3} ln ∣ x + 6 ∣

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

Add a constant to the solution

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

\int sin (θ) + 1 d θ

\frac{d}{d x} (e^{5 - 7 x})

\int cosh (5 x) d x

x, 1

\int_{7}^{11} \frac{2}{x ^{3}} d x

What is the integral of 1/(x^2+9x+18) ?
The integral of 1/(x^2+9x+18) is 1/3 ln|x+3|-1/3 ln|x+6|+C