What is the integral of (x+3)/(x^2+11x+18) ?

The integral of (x+3)/(x^2+11x+18) is 1/7 ln|x+2|+6/7 ln|x+9|+C

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integral of (x+3)/(x^2+11x+18)

\int \frac{x + 3}{x ^{2} + 11 x + 18} d x

\frac{1}{7} ln ∣ x + 2 ∣ + \frac{6}{7} ln ∣ x + 9 ∣ + C

Solution steps

\int \frac{x + 3}{x ^{2} + 11 x + 18} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 5 u - 6} d u

= \int \frac{u}{u ^{2} + 5 u - 6} d u

Take the partial fraction of

\frac{u}{u ^{2} + 5 u - 6} : \frac{1}{7 ( u - 1 )} + \frac{6}{7 ( u + 6 )}

= \int \frac{1}{7 ( u - 1 )} + \frac{6}{7 ( u + 6 )} d u

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

\int \frac{1}{7 ( u - 1 )} d u = \frac{1}{7} ln ∣ u - 1 ∣

\int \frac{6}{7 ( u + 6 )} d u = \frac{6}{7} ln ∣ u + 6 ∣

= \frac{1}{7} ln ∣ u - 1 ∣ + \frac{6}{7} ln ∣ u + 6 ∣

Substitute back

u = x + 3

= \frac{1}{7} ln ∣ x + 3 - 1 ∣ + \frac{6}{7} ln ∣ x + 3 + 6 ∣

Simplify

\frac{1}{7} ln ∣ x + 3 - 1 ∣ + \frac{6}{7} ln ∣ x + 3 + 6 ∣ : \frac{1}{7} ln ∣ x + 2 ∣ + \frac{6}{7} ln ∣ x + 9 ∣

= \frac{1}{7} ln ∣ x + 2 ∣ + \frac{6}{7} ln ∣ x + 9 ∣

Add a constant to the solution

= \frac{1}{7} ln ∣ x + 2 ∣ + \frac{6}{7} ln ∣ x + 9 ∣ + C

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What is the integral of (x+3)/(x^2+11x+18) ?
The integral of (x+3)/(x^2+11x+18) is 1/7 ln|x+2|+6/7 ln|x+9|+C