integral of 1/(x^3-2x^2-4x+8)

解

\int \frac{1}{x ^{3} - 2 x ^{2} - 4 x + 8} d x

- \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )} + \frac{1}{16} ln ∣ x + 2 ∣ + C

解答ステップ

\int \frac{1}{x ^{3} - 2 x ^{2} - 4 x + 8} d x

以下の部分分数を得る:

\frac{1}{x ^{3} - 2 x ^{2} - 4 x + 8} : - \frac{1}{16 ( x - 2 )} + \frac{1}{4 ( x - 2 ) ^{2}} + \frac{1}{16 ( x + 2 )}

= \int - \frac{1}{16 ( x - 2 )} + \frac{1}{4 ( x - 2 ) ^{2}} + \frac{1}{16 ( 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 \frac{1}{16 ( x - 2 )} d x + \int \frac{1}{4 ( x - 2 ) ^{2}} d x + \int \frac{1}{16 ( x + 2 )} d x

\int \frac{1}{16 ( x - 2 )} d x = \frac{1}{16} ln ∣ x - 2 ∣

\int \frac{1}{4 ( x - 2 ) ^{2}} d x = - \frac{1}{4 ( x - 2 )}

\int \frac{1}{16 ( x + 2 )} d x = \frac{1}{16} ln ∣ x + 2 ∣

= - \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )} + \frac{1}{16} ln ∣ x + 2 ∣

定数を解答に追加する

= - \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )} + \frac{1}{16} ln ∣ x + 2 ∣ + C