What is the integral of (18x^3-36x^2+4)/(x^2-2x) ?

The integral of (18x^3-36x^2+4)/(x^2-2x) is 9x^2-4(1/2 ln|x|-1/2 ln|x-2|)+C

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integral of (18x^3-36x^2+4)/(x^2-2x)

\int \frac{18 x ^{3} - 36 x ^{2} + 4}{x ^{2} - 2 x} d x

9 x^{2} - 4 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{18 x ^{3} - 36 x ^{2} + 4}{x ^{2} - 2 x} d x

Long division

\frac{18 x ^{3} - 36 x ^{2} + 4}{x ^{2} - 2 x} : 18 x + \frac{4}{x ^{2} - 2 x}

= \int 18 x + \frac{4}{x ^{2} - 2 x} 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 18 x d x + \int \frac{4}{x ^{2} - 2 x} d x

\int 18 x d x = 9 x^{2}

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

= 9 x^{2} - 4 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

Add a constant to the solution

= 9 x^{2} - 4 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

e^{8 x}

(- 2, - 2), (- 1, 2)

\frac{\partial}{\partial x} (e^{x^{2} - y^{2} + 4 y})

\int_{1}^{4} (x^{2} - 2) d x

f (x) = 3 sin (3 x)

What is the integral of (18x^3-36x^2+4)/(x^2-2x) ?
The integral of (18x^3-36x^2+4)/(x^2-2x) is 9x^2-4(1/2 ln|x|-1/2 ln|x-2|)+C