What is the integral of (x^2-9)/(x-3) ?

The integral of (x^2-9)/(x-3) is 1/2 (x-3)^2+6x-18+C

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integral of (x^2-9)/(x-3)

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

\frac{1}{2} (x - 3)^{2} + 6 x - 18 + C

Solution steps

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

Expand

\frac{x ^{2} - 9}{x - 3} : \frac{x ^{2}}{x - 3} - \frac{9}{x - 3}

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 - 3} d x - \int \frac{9}{x - 3} d x

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

\int \frac{9}{x - 3} d x = 9 ln ∣ x - 3 ∣

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

Simplify

\frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln ∣ x - 3 ∣ - 9 ln ∣ x - 3 ∣ : \frac{1}{2} (x - 3)^{2} + 6 x - 18

= \frac{1}{2} (x - 3)^{2} + 6 x - 18

Add a constant to the solution

= \frac{1}{2} (x - 3)^{2} + 6 x - 18 + C

\frac{d}{d x} ((x^{3} + 1)^{\frac{1}{3}})

\frac{\partial}{\partial K} (- 2 K^{2} + L^{2})

\int \frac{2 x - 1}{x ^{2} - 7} d x

\frac{\partial}{\partial x} (zy)

\frac{\partial}{\partial z} (x^{3} y z^{2})

What is the integral of (x^2-9)/(x-3) ?
The integral of (x^2-9)/(x-3) is 1/2 (x-3)^2+6x-18+C