What is the integral of (1-x)/(x^2+3x+2) ?

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

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integral of (1-x)/(x^2+3x+2)

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

2 ln ∣ x + 1 ∣ - 3 ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1 - x}{x ^{2} + 3 x + 2} : \frac{2}{x + 1} - \frac{3}{x + 2}

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

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}{x + 1} d x - \int \frac{3}{x + 2} d x

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

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

= 2 ln ∣ x + 1 ∣ - 3 ln ∣ x + 2 ∣

Add a constant to the solution

= 2 ln ∣ x + 1 ∣ - 3 ln ∣ x + 2 ∣ + C

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