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

解

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

- \frac{2}{9} ln ∣ x ∣ + \frac{2}{3 x} + \frac{1}{18} (11 ln x^{2} + 3 + 2 x - 2 arctan (\frac{1}{2} (1 + x))) + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{3} - 2 x - 2}{x ^{4} + 2 x ^{3} + 3 x ^{2}} : - \frac{2}{9 x} - \frac{2}{3 x ^{2}} + \frac{11 x + 10}{9 ( x ^{2} + 2 x + 3 )}

= \int - \frac{2}{9 x} - \frac{2}{3 x ^{2}} + \frac{11 x + 10}{9 ( x ^{2} + 2 x + 3 )} d 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{2}{9 x} d x - \int \frac{2}{3 x ^{2}} d x + \int \frac{11 x + 10}{9 ( x ^{2} + 2 x + 3 )} d x

\int \frac{2}{9 x} d x = \frac{2}{9} ln ∣ x ∣

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

\int \frac{11 x + 10}{9 ( x ^{2} + 2 x + 3 )} d x = \frac{1}{18} (11 ln x^{2} + 3 + 2 x - 2 arctan (\frac{1 + x}{2}))

= - \frac{2}{9} ln ∣ x ∣ - (- \frac{2}{3 x}) + \frac{1}{18} (11 ln x^{2} + 3 + 2 x - 2 arctan (\frac{1 + x}{2}))

簡素化

- \frac{2}{9} ln ∣ x ∣ - (- \frac{2}{3 x}) + \frac{1}{18} (11 ln x^{2} + 3 + 2 x - 2 arctan (\frac{1 + x}{2})) : - \frac{2}{9} ln ∣ x ∣ + \frac{2}{3 x} + \frac{1}{18} (11 ln x^{2} + 3 + 2 x - 2 arctan (\frac{1}{2} (1 + x)))

= - \frac{2}{9} ln ∣ x ∣ + \frac{2}{3 x} + \frac{1}{18} (11 ln x^{2} + 3 + 2 x - 2 arctan (\frac{1}{2} (1 + x)))

定数を解答に追加する

= - \frac{2}{9} ln ∣ x ∣ + \frac{2}{3 x} + \frac{1}{18} (11 ln x^{2} + 3 + 2 x - 2 arctan (\frac{1}{2} (1 + x))) + C