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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

\int 2 x d x = x^{2}

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

\int \frac{3}{x - 3} d x = 3 ln ∣ x - 3 ∣

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

定数を解答に追加する

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