What is the integral of (x^2-5x+2)/(x^3-x^2-x+1) ?

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

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integral of (x^2-5x+2)/(x^3-x^2-x+1)

\int \frac{x ^{2} - 5 x + 2}{x ^{3} - x ^{2} - x + 1} d x

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

Solution steps

\int \frac{x ^{2} - 5 x + 2}{x ^{3} - x ^{2} - x + 1} d x

Expand

\frac{x ^{2} - 5 x + 2}{x ^{3} - x ^{2} - x + 1} : \frac{x ^{2}}{x ^{3} - x ^{2} - x + 1} - \frac{5 x}{x ^{3} - x ^{2} - x + 1} + \frac{2}{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{x ^{2}}{x ^{3} - x ^{2} - x + 1} d x - \int \frac{5 x}{x ^{3} - x ^{2} - x + 1} d x + \int \frac{2}{x ^{3} - x ^{2} - x + 1} d x

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

\int \frac{5 x}{x ^{3} - x ^{2} - x + 1} d x = 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - \frac{1}{4} ln ∣ x + 1 ∣)

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

= \frac{3}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - \frac{1}{4} ln ∣ x + 1 ∣) - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{x - 1} + \frac{1}{2} ln ∣ x + 1 ∣

Simplify

\frac{3}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - \frac{1}{4} ln ∣ x + 1 ∣) - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{x - 1} + \frac{1}{2} ln ∣ x + 1 ∣ : - ln ∣ x - 1 ∣ + 2 ln ∣ x + 1 ∣ + \frac{1}{x - 1}

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

Add a constant to the solution

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

\int - \frac{24}{x ^{25}} d x

\int r^{2} + 1 d r

\int \frac{- 3}{( x - 1 ) ^{2}} d x

\int - \frac{1}{sin ^{2} ( x )} d x

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