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

The integral of (x^2+3x+2)/(x+4) is -5x+6ln|x+4|+1/2 (x+4)^2-20+C

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

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

- 5 x + 6 ln ∣ x + 4 ∣ + \frac{1}{2} (x + 4)^{2} - 20 + C

Solution steps

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

Expand

\frac{x ^{2} + 3 x + 2}{x + 4} : \frac{x ^{2}}{x + 4} + \frac{3 x}{x + 4} + \frac{2}{x + 4}

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

\int \frac{x ^{2}}{x + 4} d x = \frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣

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

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

= \frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣ + 3 (x + 4 - 4 ln ∣ x + 4 ∣) + 2 ln ∣ x + 4 ∣

Simplify

\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣ + 3 (x + 4 - 4 ln ∣ x + 4 ∣) + 2 ln ∣ x + 4 ∣ : - 5 x + 6 ln ∣ x + 4 ∣ + \frac{1}{2} (x + 4)^{2} - 20

= - 5 x + 6 ln ∣ x + 4 ∣ + \frac{1}{2} (x + 4)^{2} - 20

Add a constant to the solution

= - 5 x + 6 ln ∣ x + 4 ∣ + \frac{1}{2} (x + 4)^{2} - 20 + C

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What is the integral of (x^2+3x+2)/(x+4) ?
The integral of (x^2+3x+2)/(x+4) is -5x+6ln|x+4|+1/2 (x+4)^2-20+C