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

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

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

\int \frac{( x ^{3} - 2 x ^{2} - 4 )}{( x ^{3} - 2 x ^{2} )} d x

x + ln ∣ x ∣ - \frac{2}{x} - ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3} - 2 x ^{2} - 4}{x ^{3} - 2 x ^{2}} : 1 + \frac{1}{x} + \frac{2}{x ^{2}} - \frac{1}{x - 2}

= \int 1 + \frac{1}{x} + \frac{2}{x ^{2}} - \frac{1}{x - 2} 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 1 d x + \int \frac{1}{x} d x + \int \frac{2}{x ^{2}} d x - \int \frac{1}{x - 2} d x

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{2}{x ^{2}} d x = - \frac{2}{x}

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

= x + ln ∣ x ∣ - \frac{2}{x} - ln ∣ x - 2 ∣

Add a constant to the solution

= x + ln ∣ x ∣ - \frac{2}{x} - ln ∣ x - 2 ∣ + C

\int \frac{x ^{4}}{4} d x

\frac{d y}{d x} = y (x y^{5} - 1)

x y^{^{'}} + y = 0

arccos (x) + x

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