What is the integral of (11-2x)/(x^2+9x+14) ?

The integral of (11-2x)/(x^2+9x+14) is

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integral of (11-2x)/(x^2+9x+14)

\int \frac{11 - 2 x}{x ^{2} + 9 x + 14} d x

\frac{11}{5} ln ∣ x + 2 ∣ - \frac{11}{5} ln ∣ x + 7 ∣ - ln 4 x^{2} + 36 x + 56 - \frac{9}{5} ln \frac{2}{5} x + \frac{14}{5} + \frac{9}{5} ln \frac{2}{5} x + \frac{4}{5} + C

Solution steps

\int \frac{11 - 2 x}{x ^{2} + 9 x + 14} d x

Expand

\frac{11 - 2 x}{x ^{2} + 9 x + 14} : \frac{11}{x ^{2} + 9 x + 14} - \frac{2 x}{x ^{2} + 9 x + 14}

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{11}{x ^{2} + 9 x + 14} d x - \int \frac{2 x}{x ^{2} + 9 x + 14} d x

\int \frac{11}{x ^{2} + 9 x + 14} d x = 11 (\frac{1}{5} ln ∣ x + 2 ∣ - \frac{1}{5} ln ∣ x + 7 ∣)

\int \frac{2 x}{x ^{2} + 9 x + 14} d x = ln 4 x^{2} + 36 x + 56 + \frac{9}{5} (ln \frac{2}{5} x + \frac{14}{5} - ln \frac{2}{5} x + \frac{4}{5})

= 11 (\frac{1}{5} ln ∣ x + 2 ∣ - \frac{1}{5} ln ∣ x + 7 ∣) - (ln 4 x^{2} + 36 x + 56 + \frac{9}{5} (ln \frac{2}{5} x + \frac{14}{5} - ln \frac{2}{5} x + \frac{4}{5}))

Simplify

11 (\frac{1}{5} ln ∣ x + 2 ∣ - \frac{1}{5} ln ∣ x + 7 ∣) - (ln 4 x^{2} + 36 x + 56 + \frac{9}{5} (ln \frac{2}{5} x + \frac{14}{5} - ln \frac{2}{5} x + \frac{4}{5})) : \frac{11}{5} ln ∣ x + 2 ∣ - \frac{11}{5} ln ∣ x + 7 ∣ - ln 4 x^{2} + 36 x + 56 - \frac{9}{5} ln \frac{2}{5} x + \frac{14}{5} + \frac{9}{5} ln \frac{2}{5} x + \frac{4}{5}

= \frac{11}{5} ln ∣ x + 2 ∣ - \frac{11}{5} ln ∣ x + 7 ∣ - ln 4 x^{2} + 36 x + 56 - \frac{9}{5} ln \frac{2}{5} x + \frac{14}{5} + \frac{9}{5} ln \frac{2}{5} x + \frac{4}{5}

Add a constant to the solution

= \frac{11}{5} ln ∣ x + 2 ∣ - \frac{11}{5} ln ∣ x + 7 ∣ - ln 4 x^{2} + 36 x + 56 - \frac{9}{5} ln \frac{2}{5} x + \frac{14}{5} + \frac{9}{5} ln \frac{2}{5} x + \frac{4}{5} + C

What is the integral of (11-2x)/(x^2+9x+14) ?
The integral of (11-2x)/(x^2+9x+14) is