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

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

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

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

- \frac{x}{x ^{4} + 2 x ^{2} + 1} + \frac{x}{2 ( x ^{2} + 1 )} - \frac{1}{2} ln x^{2} + 1 - arctan (x) + \frac{1}{8} sin (4 arctan (x)) + ln ∣ x ∣ + \frac{1}{2 ( x ^{2} + 1 )} - \frac{1}{x ^{2} + 1} + C

Solution steps

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

Expand

\frac{1 - x + 2 x ^{2} - x ^{3}}{x ^{5} + 2 x ^{3} + x} : \frac{1}{x ^{5} + 2 x ^{3} + x} - \frac{x}{x ^{5} + 2 x ^{3} + x} + \frac{2 x ^{2}}{x ^{5} + 2 x ^{3} + x} - \frac{x ^{3}}{x ^{5} + 2 x ^{3} + 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{1}{x ^{5} + 2 x ^{3} + x} d x - \int \frac{x}{x ^{5} + 2 x ^{3} + x} d x + \int \frac{2 x ^{2}}{x ^{5} + 2 x ^{3} + x} d x - \int \frac{x ^{3}}{x ^{5} + 2 x ^{3} + x} d x

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

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

\int \frac{2 x ^{2}}{x ^{5} + 2 x ^{3} + x} d x = - \frac{1}{x ^{2} + 1}

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

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

Simplify

ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 + \frac{1}{2 ( x ^{2} + 1 )} - (\frac{x}{x ^{4} + 2 x ^{2} + 1} + \frac{1}{2} (arctan (x) - \frac{1}{4} sin (4 arctan (x)))) - \frac{1}{x ^{2} + 1} - (\frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )}) : - \frac{x}{x ^{4} + 2 x ^{2} + 1} + \frac{x}{2 ( x ^{2} + 1 )} - \frac{1}{2} ln x^{2} + 1 - arctan (x) + \frac{1}{8} sin (4 arctan (x)) + ln ∣ x ∣ + \frac{1}{2 ( x ^{2} + 1 )} - \frac{1}{x ^{2} + 1}

= - \frac{x}{x ^{4} + 2 x ^{2} + 1} + \frac{x}{2 ( x ^{2} + 1 )} - \frac{1}{2} ln x^{2} + 1 - arctan (x) + \frac{1}{8} sin (4 arctan (x)) + ln ∣ x ∣ + \frac{1}{2 ( x ^{2} + 1 )} - \frac{1}{x ^{2} + 1}

Add a constant to the solution

= - \frac{x}{x ^{4} + 2 x ^{2} + 1} + \frac{x}{2 ( x ^{2} + 1 )} - \frac{1}{2} ln x^{2} + 1 - arctan (x) + \frac{1}{8} sin (4 arctan (x)) + ln ∣ x ∣ + \frac{1}{2 ( x ^{2} + 1 )} - \frac{1}{x ^{2} + 1} + C

\int_{0}^{1} 6 (1 + x)^{8} d x

\int_{2}^{3} \frac{10}{x ^{2} - 1} d x

y^{^{'}} = 10 y^{2} sin (x)

\int_{0}^{8} π (x + 1)^{2} d x

\int sin^{\frac{1}{2}} (2 z) cos^{3} (2 z) d z

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