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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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