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

The integral of (x^3+3)/(x^2+3x+2) is (x^2)/2-3x-ln|x+1|+8ln|x+2|-6(1/2 ln|2x+4|-1/2 ln|2x+2|)+C

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

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

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

Solution steps

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

Expand

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

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

Add a constant to the solution

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

f (x) = \frac{4 x}{x ^{2} + 1}, (- 1, - 2)

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

(3 \frac{y}{x} - 8) d x + 3 d y = 0

y = \frac{1}{sin ^{2} ( x )}

\frac{d y}{d x} = 7 x cos (x^{2})

What is the integral of (x^3+3)/(x^2+3x+2) ?
The integral of (x^3+3)/(x^2+3x+2) is (x^2)/2-3x-ln|x+1|+8ln|x+2|-6(1/2 ln|2x+4|-1/2 ln|2x+2|)+C