What is the integral of (18x^3-36x^2+5)/(x^2-2x) ?

The integral of (18x^3-36x^2+5)/(x^2-2x) is 9x^2-5(1/2 ln|x|-1/2 ln|x-2|)+C

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integral of (18x^3-36x^2+5)/(x^2-2x)

\int \frac{18 x ^{3} - 36 x ^{2} + 5}{x ^{2} - 2 x} d x

9 x^{2} - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{18 x ^{3} - 36 x ^{2} + 5}{x ^{2} - 2 x} d x

Long division

\frac{18 x ^{3} - 36 x ^{2} + 5}{x ^{2} - 2 x} : 18 x + \frac{5}{x ^{2} - 2 x}

= \int 18 x + \frac{5}{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{5}{x ^{2} - 2 x} d x

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

\int \frac{5}{x ^{2} - 2 x} d x = - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

= 9 x^{2} - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

Add a constant to the solution

= 9 x^{2} - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

\frac{d ^{3}}{d x ^{3}} (x e^{x})

\int x (3 x + 8)^{5} d x

\int 4 tan (x) sec^{5} (x) d x

\frac{d}{d x} (- 5 x^{4} + 4 x^{3} - 20 x - 20)

n = 1 \sum \infty cos (\frac{10}{n})

What is the integral of (18x^3-36x^2+5)/(x^2-2x) ?
The integral of (18x^3-36x^2+5)/(x^2-2x) is 9x^2-5(1/2 ln|x|-1/2 ln|x-2|)+C