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

The integral of (x^3+x+1)/(x^2+3x+2) is (x^2)/2-3x-2ln|x+1|+10ln|x+2|-ln|2x+4|+ln|2x+2|+C

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

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

\frac{x ^{2}}{2} - 3 x - 2 ln ∣ x + 1 ∣ + 10 ln ∣ x + 2 ∣ - ln ∣ 2 x + 4 ∣ + ln ∣ 2 x + 2 ∣ + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

= \frac{x ^{2}}{2} - 3 x - 2 ln ∣ x + 1 ∣ + 10 ln ∣ x + 2 ∣ - ln ∣ 2 x + 4 ∣ + ln ∣ 2 x + 2 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 3 x - 2 ln ∣ x + 1 ∣ + 10 ln ∣ x + 2 ∣ - ln ∣ 2 x + 4 ∣ + ln ∣ 2 x + 2 ∣ + C

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What is the integral of (x^3+x+1)/(x^2+3x+2) ?
The integral of (x^3+x+1)/(x^2+3x+2) is (x^2)/2-3x-2ln|x+1|+10ln|x+2|-ln|2x+4|+ln|2x+2|+C