What is the integral of ((x-4))/(x^2+3x+2) ?

The integral of ((x-4))/(x^2+3x+2) is -ln|x+1|+2ln|x+2|+8(1/2 ln|2x+4|-1/2 ln|2x+2|)+C

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integral of ((x-4))/(x^2+3x+2)

\int \frac{( x - 4 )}{x ^{2} + 3 x + 2} d x

- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ + 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣) + C

Solution steps

\int \frac{x - 4}{x ^{2} + 3 x + 2} d x

Expand

\frac{x - 4}{x ^{2} + 3 x + 2} : \frac{x}{x ^{2} + 3 x + 2} - \frac{4}{x ^{2} + 3 x + 2}

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} + 3 x + 2} d x - \int \frac{4}{x ^{2} + 3 x + 2} d x

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

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

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

Simplify

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

Add a constant to the solution

= - ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ + 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣) + C

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What is the integral of ((x-4))/(x^2+3x+2) ?
The integral of ((x-4))/(x^2+3x+2) is -ln|x+1|+2ln|x+2|+8(1/2 ln|2x+4|-1/2 ln|2x+2|)+C