What is the integral of (3x^2-6x+7)/(x^3-x^2-x+1) ?

The integral of (3x^2-6x+7)/(x^3-x^2-x+1) is -ln|x-1|+4ln|x+1|-2/(x-1)+C

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integral of (3x^2-6x+7)/(x^3-x^2-x+1)

\int \frac{3 x ^{2} - 6 x + 7}{x ^{3} - x ^{2} - x + 1} d x

- ln ∣ x - 1 ∣ + 4 ln ∣ x + 1 ∣ - \frac{2}{x - 1} + C

Solution steps

\int \frac{3 x ^{2} - 6 x + 7}{x ^{3} - x ^{2} - x + 1} d x

Expand

\frac{3 x ^{2} - 6 x + 7}{x ^{3} - x ^{2} - x + 1} : \frac{3 x ^{2}}{x ^{3} - x ^{2} - x + 1} - \frac{6 x}{x ^{3} - x ^{2} - x + 1} + \frac{7}{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{3 x ^{2}}{x ^{3} - x ^{2} - x + 1} d x - \int \frac{6 x}{x ^{3} - x ^{2} - x + 1} d x + \int \frac{7}{x ^{3} - x ^{2} - x + 1} d x

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

\int \frac{6 x}{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 ∣)

\int \frac{7}{x ^{3} - x ^{2} - x + 1} d x = 7 (- \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣)

= 3 (\frac{3}{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 ∣) + 7 (- \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣)

Simplify

3 (\frac{3}{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 ∣) + 7 (- \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣) : - ln ∣ x - 1 ∣ + 4 ln ∣ x + 1 ∣ - \frac{2}{x - 1}

= - ln ∣ x - 1 ∣ + 4 ln ∣ x + 1 ∣ - \frac{2}{x - 1}

Add a constant to the solution

= - ln ∣ x - 1 ∣ + 4 ln ∣ x + 1 ∣ - \frac{2}{x - 1} + C

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What is the integral of (3x^2-6x+7)/(x^3-x^2-x+1) ?
The integral of (3x^2-6x+7)/(x^3-x^2-x+1) is -ln|x-1|+4ln|x+1|-2/(x-1)+C