What is the integral of (4x^3-5x^2)/((x-2)(x+1)) ?

The integral of (4x^3-5x^2)/((x-2)(x+1)) is

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integral of (4x^3-5x^2)/((x-2)(x+1))

\int \frac{4 x ^{3} - 5 x ^{2}}{( x - 2 ) ( x + 1 )} d x

2 x^{2} - x + 7 (\frac{1}{3} ln ∣ x + 1 ∣ + \frac{2}{3} ln ∣ x - 2 ∣) + \frac{2}{3} (ln \frac{2}{3} x + \frac{2}{3} - ln \frac{2}{3} x - \frac{4}{3}) + C

Solution steps

\int \frac{4 x ^{3} - 5 x ^{2}}{( x - 2 ) ( x + 1 )} d x

Long division

\frac{4 x ^{3} - 5 x ^{2}}{( x - 2 ) ( x + 1 )} : 4 x - 1 + \frac{7 x - 2}{x ^{2} - x - 2}

= \int 4 x - 1 + \frac{7 x - 2}{x ^{2} - 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 4 x d x - \int 1 d x + \int \frac{7 x - 2}{x ^{2} - x - 2} d x

\int 4 x d x = 2 x^{2}

\int 1 d x = x

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

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

Add a constant to the solution

= 2 x^{2} - x + 7 (\frac{1}{3} ln ∣ x + 1 ∣ + \frac{2}{3} ln ∣ x - 2 ∣) + \frac{2}{3} (ln \frac{2}{3} x + \frac{2}{3} - ln \frac{2}{3} x - \frac{4}{3}) + C

6 x, \frac{2}{3} x, 40 - x^{2}

\frac{d}{d x} (6 cos (x) sin (x))

\frac{\partial}{\partial x} (e^{5 y})

\int \frac{1}{x ^{2} \cdot 4 - x ^{2}} d x

x \to 0 - lim (\frac{5}{x} - \frac{5}{∣ x ∣})

What is the integral of (4x^3-5x^2)/((x-2)(x+1)) ?
The integral of (4x^3-5x^2)/((x-2)(x+1)) is