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

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

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

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

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

Solution steps

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

Expand

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

Simplify

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

\frac{d}{d x} (\frac{1}{x ^{3} - 2 x})

n \to \infty lim (- 1^{n})

\int_{0}^{π} 4 sin (x) + sin (4 x) d x

\frac{\partial}{\partial y} (1 - \frac{3}{x} + y)

x \to 0 lim (\frac{x ^{3} - 8}{x - 2})

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