integral of (2x^3+x-1)/(x^3+x^2-4x-4)

解

\int \frac{2 x ^{3} + x - 1}{x ^{3} + x ^{2} - 4 x - 4} d x

2 x + \frac{4}{3} ln ∣ x + 1 ∣ - \frac{19}{4} ln ∣ x + 2 ∣ + \frac{17}{12} ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{2 x ^{3} + x - 1}{x ^{3} + x ^{2} - 4 x - 4} d x

以下の部分分数を得る:

\frac{2 x ^{3} + x - 1}{x ^{3} + x ^{2} - 4 x - 4} : 2 + \frac{4}{3 ( x + 1 )} - \frac{19}{4 ( x + 2 )} + \frac{17}{12 ( x - 2 )}

= \int 2 + \frac{4}{3 ( x + 1 )} - \frac{19}{4 ( x + 2 )} + \frac{17}{12 ( 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 2 d x + \int \frac{4}{3 ( x + 1 )} d x - \int \frac{19}{4 ( x + 2 )} d x + \int \frac{17}{12 ( x - 2 )} d x

\int 2 d x = 2 x

\int \frac{4}{3 ( x + 1 )} d x = \frac{4}{3} ln ∣ x + 1 ∣

\int \frac{19}{4 ( x + 2 )} d x = \frac{19}{4} ln ∣ x + 2 ∣

\int \frac{17}{12 ( x - 2 )} d x = \frac{17}{12} ln ∣ x - 2 ∣

= 2 x + \frac{4}{3} ln ∣ x + 1 ∣ - \frac{19}{4} ln ∣ x + 2 ∣ + \frac{17}{12} ln ∣ x - 2 ∣

定数を解答に追加する

= 2 x + \frac{4}{3} ln ∣ x + 1 ∣ - \frac{19}{4} ln ∣ x + 2 ∣ + \frac{17}{12} ln ∣ x - 2 ∣ + C