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

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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

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t a y l or x, 1

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

\frac{\partial}{\partial y} ((9 x - 4 y) (12 x + 2 y))

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