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

解

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

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

解答ステップ

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

拡張

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

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

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

\int \frac{3}{( x - 1 ) ( x ^{2} - 1 )} d x = 3 (- \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣)

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

簡素化

\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - \frac{1}{4} ln ∣ x + 1 ∣ - 3 (- \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣) : ln ∣ x - 1 ∣ - ln ∣ x + 1 ∣ + \frac{1}{x - 1}

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

定数を解答に追加する

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