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

The integral of (3x^2-2x+4)/(x+2) is -14x+20ln|x+2|+3/2 (x+2)^2-28+C

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

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

- 14 x + 20 ln ∣ x + 2 ∣ + \frac{3}{2} (x + 2)^{2} - 28 + C

Solution steps

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

Expand

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

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

\int \frac{2 x}{x + 2} d x = 2 (x + 2 - 2 ln ∣ x + 2 ∣)

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

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

Simplify

3 (\frac{( x + 2 ) ^{2}}{2} - 4 (x + 2) + 4 ln ∣ x + 2 ∣) - 2 (x + 2 - 2 ln ∣ x + 2 ∣) + 4 ln ∣ x + 2 ∣ : - 14 x + 20 ln ∣ x + 2 ∣ + \frac{3}{2} (x + 2)^{2} - 28

= - 14 x + 20 ln ∣ x + 2 ∣ + \frac{3}{2} (x + 2)^{2} - 28

Add a constant to the solution

= - 14 x + 20 ln ∣ x + 2 ∣ + \frac{3}{2} (x + 2)^{2} - 28 + C

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What is the integral of (3x^2-2x+4)/(x+2) ?
The integral of (3x^2-2x+4)/(x+2) is -14x+20ln|x+2|+3/2 (x+2)^2-28+C