What is the integral of (6x^2+10x+2)/(x^3+3x^2+2x) ?

The integral of (6x^2+10x+2)/(x^3+3x^2+2x) is 2ln|x+1|+3ln|x+2|+ln|x|+C

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integral of (6x^2+10x+2)/(x^3+3x^2+2x)

\int \frac{6 x ^{2} + 10 x + 2}{x ^{3} + 3 x ^{2} + 2 x} d x

2 ln ∣ x + 1 ∣ + 3 ln ∣ x + 2 ∣ + ln ∣ x ∣ + C

Solution steps

\int \frac{6 x ^{2} + 10 x + 2}{x ^{3} + 3 x ^{2} + 2 x} d x

Expand

\frac{6 x ^{2} + 10 x + 2}{x ^{3} + 3 x ^{2} + 2 x} : \frac{6 x ^{2}}{x ^{3} + 3 x ^{2} + 2 x} + \frac{10 x}{x ^{3} + 3 x ^{2} + 2 x} + \frac{2}{x ^{3} + 3 x ^{2} + 2 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{6 x ^{2}}{x ^{3} + 3 x ^{2} + 2 x} d x + \int \frac{10 x}{x ^{3} + 3 x ^{2} + 2 x} d x + \int \frac{2}{x ^{3} + 3 x ^{2} + 2 x} d x

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

\int \frac{10 x}{x ^{3} + 3 x ^{2} + 2 x} d x = 10 (ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣)

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

= 6 (- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣) + 10 (ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣) + ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣

Simplify

6 (- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣) + 10 (ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣) + ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ : 2 ln ∣ x + 1 ∣ + 3 ln ∣ x + 2 ∣ + ln ∣ x ∣

= 2 ln ∣ x + 1 ∣ + 3 ln ∣ x + 2 ∣ + ln ∣ x ∣

Add a constant to the solution

= 2 ln ∣ x + 1 ∣ + 3 ln ∣ x + 2 ∣ + ln ∣ x ∣ + C

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What is the integral of (6x^2+10x+2)/(x^3+3x^2+2x) ?
The integral of (6x^2+10x+2)/(x^3+3x^2+2x) is 2ln|x+1|+3ln|x+2|+ln|x|+C