What is the integral of (x+4)/(x^2+11x+18) ?

The integral of (x+4)/(x^2+11x+18) is 2/7 ln|x+2|+5/7 ln|x+9|+C

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integral of (x+4)/(x^2+11x+18)

\int \frac{x + 4}{x ^{2} + 11 x + 18} d x

\frac{2}{7} ln ∣ x + 2 ∣ + \frac{5}{7} ln ∣ x + 9 ∣ + C

Solution steps

\int \frac{x + 4}{x ^{2} + 11 x + 18} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 3 u - 10} d u

= \int \frac{u}{u ^{2} + 3 u - 10} d u

Take the partial fraction of

\frac{u}{u ^{2} + 3 u - 10} : \frac{2}{7 ( u - 2 )} + \frac{5}{7 ( u + 5 )}

= \int \frac{2}{7 ( u - 2 )} + \frac{5}{7 ( u + 5 )} 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{2}{7 ( u - 2 )} d u + \int \frac{5}{7 ( u + 5 )} d u

\int \frac{2}{7 ( u - 2 )} d u = \frac{2}{7} ln ∣ u - 2 ∣

\int \frac{5}{7 ( u + 5 )} d u = \frac{5}{7} ln ∣ u + 5 ∣

= \frac{2}{7} ln ∣ u - 2 ∣ + \frac{5}{7} ln ∣ u + 5 ∣

Substitute back

u = x + 4

= \frac{2}{7} ln ∣ x + 4 - 2 ∣ + \frac{5}{7} ln ∣ x + 4 + 5 ∣

Simplify

\frac{2}{7} ln ∣ x + 4 - 2 ∣ + \frac{5}{7} ln ∣ x + 4 + 5 ∣ : \frac{2}{7} ln ∣ x + 2 ∣ + \frac{5}{7} ln ∣ x + 9 ∣

= \frac{2}{7} ln ∣ x + 2 ∣ + \frac{5}{7} ln ∣ x + 9 ∣

Add a constant to the solution

= \frac{2}{7} ln ∣ x + 2 ∣ + \frac{5}{7} ln ∣ x + 9 ∣ + C

\int 4 - x^{2} - y^{2} d x

\int x^{2} ln (1 + x) d x

\int 4 x cos (4 x^{2}) d x

\frac{d y}{d x} = y - 1

\frac{d y}{d x} = \frac{y ^{3}}{3 x - 2}, y (1) = - 1

What is the integral of (x+4)/(x^2+11x+18) ?
The integral of (x+4)/(x^2+11x+18) is 2/7 ln|x+2|+5/7 ln|x+9|+C