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

The integral of (x^2-2x+1)/(4(x-1)) is 1/8 (x^2-2x+1)+C

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

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

\frac{1}{8} (x^{2} - 2 x + 1) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

\frac{x ^{2} - 2 x + 1}{x - 1} : \frac{x ^{2}}{x - 1} - \frac{2 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

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

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

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

Simplify

\frac{1}{4} (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣ - 2 (x - 1 + ln ∣ x - 1 ∣) + ln ∣ x - 1 ∣) : \frac{1}{8} (x^{2} - 2 x + 1)

= \frac{1}{8} (x^{2} - 2 x + 1)

Add a constant to the solution

= \frac{1}{8} (x^{2} - 2 x + 1) + C

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What is the integral of (x^2-2x+1)/(4(x-1)) ?
The integral of (x^2-2x+1)/(4(x-1)) is 1/8 (x^2-2x+1)+C