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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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