What is the integral of (x^3-4x+6)/(x^2+3x) ?

The integral of (x^3-4x+6)/(x^2+3x) is -6x+1/2 (x+3)^2+3ln|x+3|+2ln|x|-18+C

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integral of (x^3-4x+6)/(x^2+3x)

\int \frac{x ^{3} - 4 x + 6}{x ^{2} + 3 x} d x

- 6 x + \frac{1}{2} (x + 3)^{2} + 3 ln ∣ x + 3 ∣ + 2 ln ∣ x ∣ - 18 + C

Solution steps

\int \frac{x ^{3} - 4 x + 6}{x ^{2} + 3 x} d x

Expand

\frac{x ^{3} - 4 x + 6}{x ^{2} + 3 x} : \frac{x ^{3}}{x ^{2} + 3 x} - \frac{4 x}{x ^{2} + 3 x} + \frac{6}{x ^{2} + 3 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{x ^{3}}{x ^{2} + 3 x} d x - \int \frac{4 x}{x ^{2} + 3 x} d x + \int \frac{6}{x ^{2} + 3 x} d x

\int \frac{x ^{3}}{x ^{2} + 3 x} d x = \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣

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

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

= \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 4 ln ∣ x + 3 ∣ + 6 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣)

Simplify

\frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 4 ln ∣ x + 3 ∣ + 6 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣) : - 6 x + \frac{1}{2} (x + 3)^{2} + 3 ln ∣ x + 3 ∣ + 2 ln ∣ x ∣ - 18

= - 6 x + \frac{1}{2} (x + 3)^{2} + 3 ln ∣ x + 3 ∣ + 2 ln ∣ x ∣ - 18

Add a constant to the solution

= - 6 x + \frac{1}{2} (x + 3)^{2} + 3 ln ∣ x + 3 ∣ + 2 ln ∣ x ∣ - 18 + C

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What is the integral of (x^3-4x+6)/(x^2+3x) ?
The integral of (x^3-4x+6)/(x^2+3x) is -6x+1/2 (x+3)^2+3ln|x+3|+2ln|x|-18+C