What is the integral of (x^3-3x^2+2x-4)/(x-1) ?

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

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integral of (x^3-3x^2+2x-4)/(x-1)

\int \frac{x ^{3} - 3 x ^{2} + 2 x - 4}{x - 1} d x

- x + \frac{1}{3} (x - 1)^{3} - 4 ln ∣ x - 1 ∣ + 1 + C

Solution steps

\int \frac{x ^{3} - 3 x ^{2} + 2 x - 4}{x - 1} d x

Expand

\frac{x ^{3} - 3 x ^{2} + 2 x - 4}{x - 1} : \frac{x ^{3}}{x - 1} - \frac{3 x ^{2}}{x - 1} + \frac{2 x}{x - 1} - \frac{4}{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 ^{3}}{x - 1} d x - \int \frac{3 x ^{2}}{x - 1} d x + \int \frac{2 x}{x - 1} d x - \int \frac{4}{x - 1} d x

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

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

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

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

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

Simplify

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

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

Add a constant to the solution

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

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