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

解

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

\frac{3}{2} ln x^{2} + 1 - \frac{x ^{3}}{x ^{2} + 1} - \frac{x}{x ^{2} + 1} - ln ∣ x ∣ + C

解答ステップ

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

拡張

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

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

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

\int \frac{x}{x ^{3} + x} d x = arctan (x)

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

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

簡素化

- (\frac{x ^{3}}{x ^{2} + 1} - arctan (x) + \frac{x}{1 + x ^{2}}) + ln x^{2} + 1 - arctan (x) - (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) : \frac{3}{2} ln x^{2} + 1 - \frac{x ^{3}}{x ^{2} + 1} - \frac{x}{x ^{2} + 1} - ln ∣ x ∣

= \frac{3}{2} ln x^{2} + 1 - \frac{x ^{3}}{x ^{2} + 1} - \frac{x}{x ^{2} + 1} - ln ∣ x ∣

定数を解答に追加する

= \frac{3}{2} ln x^{2} + 1 - \frac{x ^{3}}{x ^{2} + 1} - \frac{x}{x ^{2} + 1} - ln ∣ x ∣ + C