integral of (x^3-1)/(4x^3-x)

解

\int \frac{x ^{3} - 1}{4 x ^{3} - x} d x

\frac{1}{4} x + ln ∣ x ∣ - \frac{9}{16} ln ∣ 2 x + 1 ∣ - \frac{7}{16} ln ∣ 2 x - 1 ∣ + C

解答ステップ

\int \frac{x ^{3} - 1}{4 x ^{3} - x} d x

以下の部分分数を得る:

\frac{x ^{3} - 1}{4 x ^{3} - x} : \frac{1}{4} + \frac{1}{x} - \frac{9}{8 ( 2 x + 1 )} - \frac{7}{8 ( 2 x - 1 )}

= \int \frac{1}{4} + \frac{1}{x} - \frac{9}{8 ( 2 x + 1 )} - \frac{7}{8 ( 2 x - 1 )} 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}{4} d x + \int \frac{1}{x} d x - \int \frac{9}{8 ( 2 x + 1 )} d x - \int \frac{7}{8 ( 2 x - 1 )} d x

\int \frac{1}{4} d x = \frac{1}{4} x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{9}{8 ( 2 x + 1 )} d x = \frac{9}{16} ln ∣ 2 x + 1 ∣

\int \frac{7}{8 ( 2 x - 1 )} d x = \frac{7}{16} ln ∣ 2 x - 1 ∣

= \frac{1}{4} x + ln ∣ x ∣ - \frac{9}{16} ln ∣ 2 x + 1 ∣ - \frac{7}{16} ln ∣ 2 x - 1 ∣

定数を解答に追加する

= \frac{1}{4} x + ln ∣ x ∣ - \frac{9}{16} ln ∣ 2 x + 1 ∣ - \frac{7}{16} ln ∣ 2 x - 1 ∣ + C