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

The integral of (x^2+1)/(x^2-4x+4) is (x^3)/(x^2-4x+4)-(2x^2)/((x-2)^2)+4ln|x-2|-9/(x-2)+C

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

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

\frac{x ^{3}}{x ^{2} - 4 x + 4} - \frac{2 x ^{2}}{( x - 2 ) ^{2}} + 4 ln ∣ x - 2 ∣ - \frac{9}{x - 2} + C

Solution steps

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

Expand

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

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

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

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

Simplify

\frac{x ^{3}}{x ^{2} - 4 x + 4} + 4 (- \frac{x ^{2}}{2 ( x - 2 ) ^{2}} + ln ∣ x - 2 ∣ - \frac{2}{x - 2}) - \frac{1}{x - 2} : \frac{x ^{3}}{x ^{2} - 4 x + 4} - \frac{2 x ^{2}}{( x - 2 ) ^{2}} + 4 ln ∣ x - 2 ∣ - \frac{9}{x - 2}

= \frac{x ^{3}}{x ^{2} - 4 x + 4} - \frac{2 x ^{2}}{( x - 2 ) ^{2}} + 4 ln ∣ x - 2 ∣ - \frac{9}{x - 2}

Add a constant to the solution

= \frac{x ^{3}}{x ^{2} - 4 x + 4} - \frac{2 x ^{2}}{( x - 2 ) ^{2}} + 4 ln ∣ x - 2 ∣ - \frac{9}{x - 2} + C

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What is the integral of (x^2+1)/(x^2-4x+4) ?
The integral of (x^2+1)/(x^2-4x+4) is (x^3)/(x^2-4x+4)-(2x^2)/((x-2)^2)+4ln|x-2|-9/(x-2)+C