What is the integral of (6x^2-20x-5)/(x^3+2x^2+x) ?

The integral of (6x^2-20x-5)/(x^3+2x^2+x) is 11ln|x+1|+(21)/(x+1)-5ln|x|+C

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integral of (6x^2-20x-5)/(x^3+2x^2+x)

\int \frac{6 x ^{2} - 20 x - 5}{x ^{3} + 2 x ^{2} + x} d x

11 ln ∣ x + 1 ∣ + \frac{21}{x + 1} - 5 ln ∣ x ∣ + C

Solution steps

\int \frac{6 x ^{2} - 20 x - 5}{x ^{3} + 2 x ^{2} + x} d x

Expand

\frac{6 x ^{2} - 20 x - 5}{x ^{3} + 2 x ^{2} + x} : \frac{6 x ^{2}}{x ^{3} + 2 x ^{2} + x} - \frac{20 x}{x ^{3} + 2 x ^{2} + x} - \frac{5}{x ^{3} + 2 x ^{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} + 2 x ^{2} + x} d x - \int \frac{20 x}{x ^{3} + 2 x ^{2} + x} d x - \int \frac{5}{x ^{3} + 2 x ^{2} + x} d x

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

\int \frac{20 x}{x ^{3} + 2 x ^{2} + x} d x = - \frac{20}{x + 1}

\int \frac{5}{x ^{3} + 2 x ^{2} + x} d x = 5 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

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

Simplify

6 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - (- \frac{20}{x + 1}) - 5 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) : 11 ln ∣ x + 1 ∣ + \frac{21}{x + 1} - 5 ln ∣ x ∣

= 11 ln ∣ x + 1 ∣ + \frac{21}{x + 1} - 5 ln ∣ x ∣

Add a constant to the solution

= 11 ln ∣ x + 1 ∣ + \frac{21}{x + 1} - 5 ln ∣ x ∣ + C

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What is the integral of (6x^2-20x-5)/(x^3+2x^2+x) ?
The integral of (6x^2-20x-5)/(x^3+2x^2+x) is 11ln|x+1|+(21)/(x+1)-5ln|x|+C