integral of 1/((x^2-x-2)(x-2))

解

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

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

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( x ^{2} - x - 2 ) ( x - 2 )} : \frac{1}{9 ( x + 1 )} - \frac{1}{9 ( x - 2 )} + \frac{1}{3 ( x - 2 ) ^{2}}

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

\int \frac{1}{9 ( x + 1 )} d x = \frac{1}{9} ln ∣ x + 1 ∣

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

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

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

定数を解答に追加する

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