What is the integral of (6x^3-12x^2+4)/(x^2-2x) ?

The integral of (6x^3-12x^2+4)/(x^2-2x) is 3x^2+2ln|x-2|-2ln|x|-12+C

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integral of (6x^3-12x^2+4)/(x^2-2x)

\int \frac{6 x ^{3} - 12 x ^{2} + 4}{x ^{2} - 2 x} d x

3 x^{2} + 2 ln ∣ x - 2 ∣ - 2 ln ∣ x ∣ - 12 + C

Solution steps

\int \frac{6 x ^{3} - 12 x ^{2} + 4}{x ^{2} - 2 x} d x

Expand

\frac{6 x ^{3} - 12 x ^{2} + 4}{x ^{2} - 2 x} : \frac{6 x ^{3}}{x ^{2} - 2 x} - \frac{12 x ^{2}}{x ^{2} - 2 x} + \frac{4}{x ^{2} - 2 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 ^{3}}{x ^{2} - 2 x} d x - \int \frac{12 x ^{2}}{x ^{2} - 2 x} d x + \int \frac{4}{x ^{2} - 2 x} d x

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

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

\int \frac{4}{x ^{2} - 2 x} d x = - 4 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

= 6 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣) - 12 (x - 2 + 2 ln ∣ x - 2 ∣) - 4 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

Simplify

6 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣) - 12 (x - 2 + 2 ln ∣ x - 2 ∣) - 4 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) : 3 x^{2} + 2 ln ∣ x - 2 ∣ - 2 ln ∣ x ∣ - 12

= 3 x^{2} + 2 ln ∣ x - 2 ∣ - 2 ln ∣ x ∣ - 12

Add a constant to the solution

= 3 x^{2} + 2 ln ∣ x - 2 ∣ - 2 ln ∣ x ∣ - 12 + C

x \to 2 π lim (\frac{x - 2 π}{sin ( x )})

\frac{d}{d x} (tan (b x))

x cos (5 x)

(y + x + \frac{y ^{2}}{x}) d x - x d y = 0

\int 6 sin (x) + 5 d x

What is the integral of (6x^3-12x^2+4)/(x^2-2x) ?
The integral of (6x^3-12x^2+4)/(x^2-2x) is 3x^2+2ln|x-2|-2ln|x|-12+C