What is the integral of (x-5)/(x^2+5x+4) ?

The integral of (x-5)/(x^2+5x+4) is

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integral of (x-5)/(x^2+5x+4)

\int \frac{x - 5}{x ^{2} + 5 x + 4} d x

2 (\frac{1}{4} ln 4 x^{2} + 20 x + 16 + \frac{5}{12} (ln \frac{8}{3} + \frac{2}{3} x - ln \frac{2}{3} + \frac{2}{3} x)) - 5 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 4 ∣) + C

Solution steps

\int \frac{x - 5}{x ^{2} + 5 x + 4} d x

Expand

\frac{x - 5}{x ^{2} + 5 x + 4} : \frac{x}{x ^{2} + 5 x + 4} - \frac{5}{x ^{2} + 5 x + 4}

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{x}{x ^{2} + 5 x + 4} d x - \int \frac{5}{x ^{2} + 5 x + 4} d x

\int \frac{x}{x ^{2} + 5 x + 4} d x = 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 16 + \frac{5}{12} (ln \frac{8}{3} + \frac{2}{3} x - ln \frac{2}{3} + \frac{2}{3} x))

\int \frac{5}{x ^{2} + 5 x + 4} d x = 5 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 4 ∣)

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 16 + \frac{5}{12} (ln \frac{8}{3} + \frac{2}{3} x - ln \frac{2}{3} + \frac{2}{3} x)) - 5 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 4 ∣)

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 16 + \frac{5}{12} (ln \frac{8}{3} + \frac{2}{3} x - ln \frac{2}{3} + \frac{2}{3} x)) - 5 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 4 ∣) + C