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

解

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

x + \frac{8}{9} ln ∣ x - 1 ∣ - \frac{8}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{2}{3 ( x - 1 )} + C

解答ステップ

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

長除法

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

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

\int 1 d x = x

\int \frac{3 x - 2}{x ^{3} - 3 x + 2} d x = \frac{8}{9} ln ∣ x - 1 ∣ - \frac{8}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{2}{3 ( x - 1 )}

= x + \frac{8}{9} ln ∣ x - 1 ∣ - \frac{8}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{2}{3 ( x - 1 )}

定数を解答に追加する

= x + \frac{8}{9} ln ∣ x - 1 ∣ - \frac{8}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{2}{3 ( x - 1 )} + C