What is the integral of (6x^2+12x+6)/(x^3+x^2-x-1) ?

The integral of (6x^2+12x+6)/(x^3+x^2-x-1) is 6ln|x-1|+C

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integral of (6x^2+12x+6)/(x^3+x^2-x-1)

\int \frac{6 x ^{2} + 12 x + 6}{x ^{3} + x ^{2} - x - 1} d x

6 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{6 x ^{2} + 12 x + 6}{x ^{3} + x ^{2} - x - 1} d x

Expand

\frac{6 x ^{2} + 12 x + 6}{x ^{3} + x ^{2} - x - 1} : \frac{6 x ^{2}}{x ^{3} + x ^{2} - x - 1} + \frac{12 x}{x ^{3} + x ^{2} - x - 1} + \frac{6}{x ^{3} + x ^{2} - x - 1}

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} + x ^{2} - x - 1} d x + \int \frac{12 x}{x ^{3} + x ^{2} - x - 1} d x + \int \frac{6}{x ^{3} + x ^{2} - x - 1} d x

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

\int \frac{12 x}{x ^{3} + x ^{2} - x - 1} d x = 12 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

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

= 6 (\frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 12 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 6 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

Simplify

6 (\frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 12 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 6 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) : 6 ln ∣ x - 1 ∣

= 6 ln ∣ x - 1 ∣

Add a constant to the solution

= 6 ln ∣ x - 1 ∣ + C

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What is the integral of (6x^2+12x+6)/(x^3+x^2-x-1) ?
The integral of (6x^2+12x+6)/(x^3+x^2-x-1) is 6ln|x-1|+C