integral of (x^3+8)/((x^2-1)(x-2))

解

\int \frac{x ^{3} + 8}{( x ^{2} - 1 ) ( x - 2 )} d x

x + \frac{7}{6} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{x ^{3} + 8}{( x ^{2} - 1 ) ( x - 2 )} d x

以下の部分分数を得る:

\frac{x ^{3} + 8}{( x ^{2} - 1 ) ( x - 2 )} : 1 + \frac{7}{6 ( x + 1 )} - \frac{9}{2 ( x - 1 )} + \frac{16}{3 ( x - 2 )}

= \int 1 + \frac{7}{6 ( x + 1 )} - \frac{9}{2 ( x - 1 )} + \frac{16}{3 ( 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 1 d x + \int \frac{7}{6 ( x + 1 )} d x - \int \frac{9}{2 ( x - 1 )} d x + \int \frac{16}{3 ( x - 2 )} d x

\int 1 d x = x

\int \frac{7}{6 ( x + 1 )} d x = \frac{7}{6} ln ∣ x + 1 ∣

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

\int \frac{16}{3 ( x - 2 )} d x = \frac{16}{3} ln ∣ x - 2 ∣

= x + \frac{7}{6} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣

定数を解答に追加する

= x + \frac{7}{6} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣ + C