What is the integral of (x^2-7x+12)/(x-3) ?

The integral of (x^2-7x+12)/(x-3) is -x+1/2 (x-3)^2+3+C

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integral of (x^2-7x+12)/(x-3)

\int \frac{x ^{2} - 7 x + 12}{x - 3} d x

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

Solution steps

\int \frac{x ^{2} - 7 x + 12}{x - 3} d x

Expand

\frac{x ^{2} - 7 x + 12}{x - 3} : \frac{x ^{2}}{x - 3} - \frac{7 x}{x - 3} + \frac{12}{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{7 x}{x - 3} d x + \int \frac{12}{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{7 x}{x - 3} d x = 7 (x - 3 + 3 ln ∣ x - 3 ∣)

\int \frac{12}{x - 3} d x = 12 ln ∣ x - 3 ∣

= \frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln ∣ x - 3 ∣ - 7 (x - 3 + 3 ln ∣ x - 3 ∣) + 12 ln ∣ x - 3 ∣

Simplify

\frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln ∣ x - 3 ∣ - 7 (x - 3 + 3 ln ∣ x - 3 ∣) + 12 ln ∣ x - 3 ∣ : - x + \frac{1}{2} (x - 3)^{2} + 3

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

Add a constant to the solution

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

What is the integral of (x^2-7x+12)/(x-3) ?
The integral of (x^2-7x+12)/(x-3) is -x+1/2 (x-3)^2+3+C