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

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

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integral of (x^3)/(2x^2-x-1)

\int \frac{x ^{3}}{2 x ^{2} - x - 1} d x

\frac{x ^{2}}{4} + \frac{1}{4} x + \frac{1}{24} (- 2 ln (\frac{∣ 2 x + 1 ∣}{2 ∣ x - 1 ∣}) (3 x + 1) + 2 (3 x + 1) (ln (\frac{2 ∣ 2 x + 1 ∣}{3}) - ln (\frac{4 ∣ x - 1 ∣}{3})) + ln ∣ 6 x + 3 ∣ + 8 ln ∣ 6 x - 6 ∣) + C

Solution steps

\int \frac{x ^{3}}{2 x ^{2} - x - 1} d x

Long division

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

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

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

\int \frac{1}{4} d x = \frac{1}{4} x

\int \frac{3 x + 1}{4 ( 2 x ^{2} - x - 1 )} d x = \frac{1}{24} (- 2 ln (\frac{∣ 2 x + 1 ∣}{2 ∣ x - 1 ∣}) (3 x + 1) + 2 (3 x + 1) (ln (\frac{2 ∣ 2 x + 1 ∣}{3}) - ln (\frac{4 ∣ x - 1 ∣}{3})) + ln ∣ 6 x + 3 ∣ + 8 ln ∣ 6 x - 6 ∣)

= \frac{x ^{2}}{4} + \frac{1}{4} x + \frac{1}{24} (- 2 ln (\frac{∣ 2 x + 1 ∣}{2 ∣ x - 1 ∣}) (3 x + 1) + 2 (3 x + 1) (ln (\frac{2 ∣ 2 x + 1 ∣}{3}) - ln (\frac{4 ∣ x - 1 ∣}{3})) + ln ∣ 6 x + 3 ∣ + 8 ln ∣ 6 x - 6 ∣)

Add a constant to the solution

= \frac{x ^{2}}{4} + \frac{1}{4} x + \frac{1}{24} (- 2 ln (\frac{∣ 2 x + 1 ∣}{2 ∣ x - 1 ∣}) (3 x + 1) + 2 (3 x + 1) (ln (\frac{2 ∣ 2 x + 1 ∣}{3}) - ln (\frac{4 ∣ x - 1 ∣}{3})) + ln ∣ 6 x + 3 ∣ + 8 ln ∣ 6 x - 6 ∣) + C

f (x) = 4 x - 3 x

\frac{d}{d x} (a x^{n})

\frac{d}{d x} (\frac{1}{x ^{4} + 1})

\int (8^{x} - sec^{2} (x)) d x

x^{2} y^{^{''}} - 2 x y^{^{'}} - 40 y = x^{4}