What is the integral of (5x-2)/(x^2+6x+8) ?

The integral of (5x-2)/(x^2+6x+8) is 5/2 ln|x^2+6x+8|+17/2 ln|x+4|-17/2 ln|x+2|+C

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integral of (5x-2)/(x^2+6x+8)

\int \frac{5 x - 2}{x ^{2} + 6 x + 8} d x

\frac{5}{2} ln x^{2} + 6 x + 8 + \frac{17}{2} ln ∣ x + 4 ∣ - \frac{17}{2} ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{5 x - 2}{x ^{2} + 6 x + 8} d x

Expand

\frac{5 x - 2}{x ^{2} + 6 x + 8} : \frac{5 x}{x ^{2} + 6 x + 8} - \frac{2}{x ^{2} + 6 x + 8}

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{5 x}{x ^{2} + 6 x + 8} d x - \int \frac{2}{x ^{2} + 6 x + 8} d x

\int \frac{5 x}{x ^{2} + 6 x + 8} d x = 5 (\frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣))

\int \frac{2}{x ^{2} + 6 x + 8} d x = ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣

= 5 (\frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣)) - (ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣)

Simplify

5 (\frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣)) - (ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣) : \frac{5}{2} ln x^{2} + 6 x + 8 + \frac{17}{2} ln ∣ x + 4 ∣ - \frac{17}{2} ln ∣ x + 2 ∣

= \frac{5}{2} ln x^{2} + 6 x + 8 + \frac{17}{2} ln ∣ x + 4 ∣ - \frac{17}{2} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{5}{2} ln x^{2} + 6 x + 8 + \frac{17}{2} ln ∣ x + 4 ∣ - \frac{17}{2} ln ∣ x + 2 ∣ + C

y^{^{''}} - 4 y = 6 e^{(- t)}, y (0) = 0, y^{^{'}} (0) = 0

x y^{^{'}} - y + 2 = 0

\int (2 sin (x) - 3 cos (x)) d x

20 t^{2}

x \to 17 lim (x^{2} - 3 x + 1)

What is the integral of (5x-2)/(x^2+6x+8) ?
The integral of (5x-2)/(x^2+6x+8) is 5/2 ln|x^2+6x+8|+17/2 ln|x+4|-17/2 ln|x+2|+C