integral of 1/((x^2-1)(x^2-4))

解

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

\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} : \frac{1}{6 ( x + 1 )} - \frac{1}{6 ( x - 1 )} - \frac{1}{12 ( x + 2 )} + \frac{1}{12 ( x - 2 )}

= \int \frac{1}{6 ( x + 1 )} - \frac{1}{6 ( x - 1 )} - \frac{1}{12 ( x + 2 )} + \frac{1}{12 ( 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}{6 ( x + 1 )} d x - \int \frac{1}{6 ( x - 1 )} d x - \int \frac{1}{12 ( x + 2 )} d x + \int \frac{1}{12 ( x - 2 )} d x

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

\int \frac{1}{6 ( x - 1 )} d x = \frac{1}{6} ln ∣ x - 1 ∣

\int \frac{1}{12 ( x + 2 )} d x = \frac{1}{12} ln ∣ x + 2 ∣

\int \frac{1}{12 ( x - 2 )} d x = \frac{1}{12} ln ∣ x - 2 ∣

= \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣

定数を解答に追加する

= \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣ + C