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

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

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

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

ln ∣ x - 1 ∣ - arctan (x) + C

Solution steps

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

Expand

\frac{x ^{2} - x + 2}{x ^{3} - x ^{2} + x - 1} : \frac{x ^{2}}{x ^{3} - x ^{2} + x - 1} - \frac{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{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{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x))

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

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

= \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) - (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x))) + ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 1 - arctan (x)

Simplify

\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) - (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x))) + ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 1 - arctan (x) : ln ∣ x - 1 ∣ - arctan (x)

= ln ∣ x - 1 ∣ - arctan (x)

Add a constant to the solution

= ln ∣ x - 1 ∣ - arctan (x) + C

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