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

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

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

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

- \frac{2}{7} arctan (\frac{1}{7} (2 x - 1)) + \frac{2}{7} arctan (\frac{1}{7} (2 x - 1)) + 2 ln 4 x^{2} - 4 x + 8 + 2 ln ∣ x ∣ + C

Solution steps

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

Expand

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

\int \frac{6 x ^{2}}{x ^{3} - x ^{2} + 2 x} d x = 12 (\frac{1}{4} ln 4 x^{2} - 4 x + 8 + \frac{1}{2 7} arctan (\frac{2 x - 1}{7}))

\int \frac{4 x}{x ^{3} - x ^{2} + 2 x} d x = \frac{8}{7} arctan (\frac{2 x - 1}{7})

\int \frac{4}{x ^{3} - x ^{2} + 2 x} d x = 2 ln ∣ x ∣ - ln 4 x^{2} - 4 x + 8 + \frac{2}{7} arctan (\frac{1}{7} (2 x - 1))

= 12 (\frac{1}{4} ln 4 x^{2} - 4 x + 8 + \frac{1}{2 7} arctan (\frac{2 x - 1}{7})) - \frac{8}{7} arctan (\frac{2 x - 1}{7}) + 2 ln ∣ x ∣ - ln 4 x^{2} - 4 x + 8 + \frac{2}{7} arctan (\frac{1}{7} (2 x - 1))

Simplify

12 (\frac{1}{4} ln 4 x^{2} - 4 x + 8 + \frac{1}{2 7} arctan (\frac{2 x - 1}{7})) - \frac{8}{7} arctan (\frac{2 x - 1}{7}) + 2 ln ∣ x ∣ - ln 4 x^{2} - 4 x + 8 + \frac{2}{7} arctan (\frac{1}{7} (2 x - 1)) : - \frac{2}{7} arctan (\frac{1}{7} (2 x - 1)) + \frac{2}{7} arctan (\frac{1}{7} (2 x - 1)) + 2 ln 4 x^{2} - 4 x + 8 + 2 ln ∣ x ∣

= - \frac{2}{7} arctan (\frac{1}{7} (2 x - 1)) + \frac{2}{7} arctan (\frac{1}{7} (2 x - 1)) + 2 ln 4 x^{2} - 4 x + 8 + 2 ln ∣ x ∣

Add a constant to the solution

= - \frac{2}{7} arctan (\frac{1}{7} (2 x - 1)) + \frac{2}{7} arctan (\frac{1}{7} (2 x - 1)) + 2 ln 4 x^{2} - 4 x + 8 + 2 ln ∣ x ∣ + C

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

\frac{d}{d x} (2 π)

\int e^{c o s (4 t)} sin (4 t) d t

x \to - 6 lim (\frac{x + 6}{∣ x + 6 ∣})

\frac{d}{d x} (x^{2} \cdot (sin (x))^{- 1})

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