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

解

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

\frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{12 x} + \frac{1}{4 ( x - 2 )} + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( x - 3 ) ( x ^{2} - 2 x ) ^{2}} : \frac{1}{9 ( x - 3 )} - \frac{1}{9 x} - \frac{1}{12 x ^{2}} - \frac{1}{4 ( x - 2 ) ^{2}}

= \int \frac{1}{9 ( x - 3 )} - \frac{1}{9 x} - \frac{1}{12 x ^{2}} - \frac{1}{4 ( 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 - 3 )} d x - \int \frac{1}{9 x} d x - \int \frac{1}{12 x ^{2}} d x - \int \frac{1}{4 ( x - 2 ) ^{2}} d x

\int \frac{1}{9 ( x - 3 )} d x = \frac{1}{9} ln ∣ x - 3 ∣

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

\int \frac{1}{12 x ^{2}} d x = - \frac{1}{12 x}

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

= \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x ∣ - (- \frac{1}{12 x}) - (- \frac{1}{4 ( x - 2 )})

簡素化

= \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{12 x} + \frac{1}{4 ( x - 2 )}

定数を解答に追加する

= \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{12 x} + \frac{1}{4 ( x - 2 )} + C