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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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