What is the integral of ((x^2-4x+9))/(x^2-6x+9) ?

The integral of ((x^2-4x+9))/(x^2-6x+9) is x+2ln|x-3|-6/(x-3)+C

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integral of ((x^2-4x+9))/(x^2-6x+9)

\int \frac{( x ^{2} - 4 x + 9 )}{x ^{2} - 6 x + 9} d x

x + 2 ln ∣ x - 3 ∣ - \frac{6}{x - 3} + C

Solution steps

\int \frac{x ^{2} - 4 x + 9}{x ^{2} - 6 x + 9} d x

Expand

\frac{x ^{2} - 4 x + 9}{x ^{2} - 6 x + 9} : \frac{x ^{2}}{x ^{2} - 6 x + 9} - \frac{4 x}{x ^{2} - 6 x + 9} + \frac{9}{x ^{2} - 6 x + 9}

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} - 6 x + 9} d x - \int \frac{4 x}{x ^{2} - 6 x + 9} d x + \int \frac{9}{x ^{2} - 6 x + 9} d x

\int \frac{x ^{2}}{x ^{2} - 6 x + 9} d x = 6 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + x

\int \frac{4 x}{x ^{2} - 6 x + 9} d x = 4 (ln ∣ x - 3 ∣ - \frac{3}{x - 3})

\int \frac{9}{x ^{2} - 6 x + 9} d x = - \frac{9}{x - 3}

= 6 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + x - 4 (ln ∣ x - 3 ∣ - \frac{3}{x - 3}) - \frac{9}{x - 3}

Simplify

6 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + x - 4 (ln ∣ x - 3 ∣ - \frac{3}{x - 3}) - \frac{9}{x - 3} : x + 2 ln ∣ x - 3 ∣ - \frac{6}{x - 3}

= x + 2 ln ∣ x - 3 ∣ - \frac{6}{x - 3}

Add a constant to the solution

= x + 2 ln ∣ x - 3 ∣ - \frac{6}{x - 3} + C

in v erse l a pl a ce \frac{( 3 s + 7 )}{s ^{2} + 2 s + 4}

d er i v a t i v e f (x) = - 3 x^{2}

t an g e n t x^{2} + x, at x = 3

\int \frac{1}{x ^{2} + 6 x + 18} d x

\frac{d}{d x} (\frac{7}{2} x^{2} - 5 x + 1)

What is the integral of ((x^2-4x+9))/(x^2-6x+9) ?
The integral of ((x^2-4x+9))/(x^2-6x+9) is x+2ln|x-3|-6/(x-3)+C