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

The integral of (x^4)/(x^3+x^2-x-1) is (x^2)/2-x+7/4 ln|x+1|+1/4 ln|x-1|+1/(x+1)-1/(2(x+1))+C

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

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

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

Solution steps

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

Long division

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

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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

\int \frac{s}{2 s + 3} d s

f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = 2

\int x (3 + 5 x) d x

x \to 2 lim (3 f (x) + 2 g (x) (x)^{2})

\frac{2 s ^{2} + 5 s + 5}{( s + 1 ) ^{3}}

What is the integral of (x^4)/(x^3+x^2-x-1) ?
The integral of (x^4)/(x^3+x^2-x-1) is (x^2)/2-x+7/4 ln|x+1|+1/4 ln|x-1|+1/(x+1)-1/(2(x+1))+C