integral of (2x^3-4)/(4x^3-2x)

解

\int \frac{2 x ^{3} - 4}{4 x ^{3} - 2 x} d x

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

解答ステップ

\int \frac{2 x ^{3} - 4}{4 x ^{3} - 2 x} d x

以下の部分分数を得る:

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

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

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

定数を解答に追加する

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