What is the integral of (2x^2-6x+4)/(x-3) ?

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

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integral of (2x^2-6x+4)/(x-3)

\int \frac{2 x ^{2} - 6 x + 4}{x - 3} d x

x^{2} + 4 ln ∣ x - 3 ∣ - 9 + C

Solution steps

\int \frac{2 x ^{2} - 6 x + 4}{x - 3} d x

Expand

\frac{2 x ^{2} - 6 x + 4}{x - 3} : \frac{2 x ^{2}}{x - 3} - \frac{6 x}{x - 3} + \frac{4}{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{2 x ^{2}}{x - 3} d x - \int \frac{6 x}{x - 3} d x + \int \frac{4}{x - 3} d x

\int \frac{2 x ^{2}}{x - 3} d x = x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27

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

\int \frac{4}{x - 3} d x = 4 ln ∣ x - 3 ∣

= x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27 - 6 (x - 3 + 3 ln ∣ x - 3 ∣) + 4 ln ∣ x - 3 ∣

Simplify

x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27 - 6 (x - 3 + 3 ln ∣ x - 3 ∣) + 4 ln ∣ x - 3 ∣ : x^{2} + 4 ln ∣ x - 3 ∣ - 9

= x^{2} + 4 ln ∣ x - 3 ∣ - 9

Add a constant to the solution

= x^{2} + 4 ln ∣ x - 3 ∣ - 9 + C

\int \frac{5 - 3 x}{9} d x

\int x^{5} - 1 d x

\int (3 x^{2} - 4 x + 2) e^{x} d x

\int \frac{1}{x ^{2} - 4 x - 21} d x

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

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