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

解

\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

解答ステップ

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

拡張

\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}

総和規則を適用する:

\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} )})

簡素化

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}

定数を解答に追加する

= - \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