What is the integral of (20x^3-40x^2+9)/(x^2-2x) ?

The integral of (20x^3-40x^2+9)/(x^2-2x) is 10x^2-9(1/2 ln|x|-1/2 ln|x-2|)+C

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integral of (20x^3-40x^2+9)/(x^2-2x)

\int \frac{20 x ^{3} - 40 x ^{2} + 9}{x ^{2} - 2 x} d x

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

Solution steps

\int \frac{20 x ^{3} - 40 x ^{2} + 9}{x ^{2} - 2 x} d x

Long division

\frac{20 x ^{3} - 40 x ^{2} + 9}{x ^{2} - 2 x} : 20 x + \frac{9}{x ^{2} - 2 x}

= \int 20 x + \frac{9}{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 20 x d x + \int \frac{9}{x ^{2} - 2 x} d x

\int 20 x d x = 10 x^{2}

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

= 10 x^{2} - 9 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

Add a constant to the solution

= 10 x^{2} - 9 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

\frac{\partial}{\partial z} (e^{x})

y^{3} - (6 x + 2) + 3 x y^{2} y^{^{'}} = 0

t an g e n t (x^{3} - 4 x)^{8}

\frac{\partial}{\partial x} (x^{2} + y^{2} - 1)

d er i v a t i v e f (x) = \frac{x ^{3}}{( x ^{2} + 4 ) ^{2}}

What is the integral of (20x^3-40x^2+9)/(x^2-2x) ?
The integral of (20x^3-40x^2+9)/(x^2-2x) is 10x^2-9(1/2 ln|x|-1/2 ln|x-2|)+C