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

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

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

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

6 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + x + C

Solution steps

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

\frac{x ^{2}}{x ^{2} - 6 x + 9} = \frac{6 x - 9}{x ^{2} - 6 x + 9} + 1

= \int \frac{6 x - 9}{x ^{2} - 6 x + 9} + 1 d x

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{6 x - 9}{x ^{2} - 6 x + 9} d x + \int 1 d x

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

\int 1 d x = x

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

Add a constant to the solution

= 6 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + x + C

y = \frac{∣ x ∣}{5 - x ^{2}}, (2, 2)

\frac{2 x ^{2} - 7}{8 x ^{3} + 1}

\frac{d}{d x} (\frac{x ^{2} + y ^{2} - 9}{x})

n = 1 \sum \infty \frac{- 1 ^{n + 1}}{n}

\int x^{2} cos (\frac{1}{2} x) d x

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