What is the integral of (30x^3-60x^2+4)/(x^2-2x) ?

The integral of (30x^3-60x^2+4)/(x^2-2x) is 15x^2-4(1/2 ln|x|-1/2 ln|x-2|)+C

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integral of (30x^3-60x^2+4)/(x^2-2x)

\int \frac{30 x ^{3} - 60 x ^{2} + 4}{x ^{2} - 2 x} d x

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

Solution steps

\int \frac{30 x ^{3} - 60 x ^{2} + 4}{x ^{2} - 2 x} d x

Long division

\frac{30 x ^{3} - 60 x ^{2} + 4}{x ^{2} - 2 x} : 30 x + \frac{4}{x ^{2} - 2 x}

= \int 30 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 30 x d x + \int \frac{4}{x ^{2} - 2 x} d x

\int 30 x d x = 15 x^{2}

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

= 15 x^{2} - 4 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

Add a constant to the solution

= 15 x^{2} - 4 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

\int \frac{( e ^{x} )}{( 1 + e ^{x} )} d x

n = 1 \sum \infty \frac{1 0 ^{n}}{n !}

\frac{\partial}{\partial x} (6 e^{x^{6} y})

2 y^{^{''}} - 2 y^{^{'}} + 25 y = 0

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

What is the integral of (30x^3-60x^2+4)/(x^2-2x) ?
The integral of (30x^3-60x^2+4)/(x^2-2x) is 15x^2-4(1/2 ln|x|-1/2 ln|x-2|)+C