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

The integral of (x^3-x^2-2x)/(4x-2) is (x^3)/(12)-(x^2)/(16)-9/16 x-9/32 ln|4x-2|+C

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

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

\frac{x ^{3}}{12} - \frac{x ^{2}}{16} - \frac{9}{16} x - \frac{9}{32} ln ∣ 4 x - 2 ∣ + C

Solution steps

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

Long division

\frac{x ^{3} - x ^{2} - 2 x}{4 x - 2} : \frac{x ^{2}}{4} - \frac{x}{8} - \frac{9}{16} - \frac{9}{8 ( 4 x - 2 )}

= \int \frac{x ^{2}}{4} - \frac{x}{8} - \frac{9}{16} - \frac{9}{8 ( 4 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 \frac{x ^{2}}{4} d x - \int \frac{x}{8} d x - \int \frac{9}{16} d x - \int \frac{9}{8 ( 4 x - 2 )} d x

\int \frac{x ^{2}}{4} d x = \frac{x ^{3}}{12}

\int \frac{x}{8} d x = \frac{x ^{2}}{16}

\int \frac{9}{16} d x = \frac{9}{16} x

\int \frac{9}{8 ( 4 x - 2 )} d x = \frac{9}{32} ln ∣ 4 x - 2 ∣

= \frac{x ^{3}}{12} - \frac{x ^{2}}{16} - \frac{9}{16} x - \frac{9}{32} ln ∣ 4 x - 2 ∣

Add a constant to the solution

= \frac{x ^{3}}{12} - \frac{x ^{2}}{16} - \frac{9}{16} x - \frac{9}{32} ln ∣ 4 x - 2 ∣ + C

x \to \infty + lim (\frac{lo g _{6} ( x )}{lo g _{8} ( x )})

\frac{d ^{5}}{d x ^{5}} (\frac{1}{x} + cos (2 x))

\frac{d}{d x} (12 x y)

f (x) = \frac{x + x}{x ^{2}}

f (x) = e^{(x^{2} - x)}

What is the integral of (x^3-x^2-2x)/(4x-2) ?
The integral of (x^3-x^2-2x)/(4x-2) is (x^3)/(12)-(x^2)/(16)-9/16 x-9/32 ln|4x-2|+C