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

解

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

ln ∣ x ∣ - \frac{3}{8} ln 16 x^{2} - 8 x + 8 - \frac{1}{4 7} arctan (\frac{1}{7} (4 x - 1)) - \frac{1}{4} ln ∣ x + 1 ∣ + C

解答ステップ

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

以下の部分分数を得る:

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

= \int \frac{1}{x} + \frac{- 6 x + 1}{4 ( 2 x ^{2} - x + 1 )} - \frac{1}{4 ( 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}{x} d x + \int \frac{- 6 x + 1}{4 ( 2 x ^{2} - x + 1 )} d x - \int \frac{1}{4 ( x + 1 )} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{- 6 x + 1}{4 ( 2 x ^{2} - x + 1 )} d x = \frac{1}{8 7} (- 37 ln 16 x^{2} - 8 x + 8 - 2 arctan (\frac{4 x - 1}{7}))

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

= ln ∣ x ∣ + \frac{1}{8 7} (- 37 ln 16 x^{2} - 8 x + 8 - 2 arctan (\frac{4 x - 1}{7})) - \frac{1}{4} ln ∣ x + 1 ∣

簡素化

ln ∣ x ∣ + \frac{1}{8 7} (- 37 ln 16 x^{2} - 8 x + 8 - 2 arctan (\frac{4 x - 1}{7})) - \frac{1}{4} ln ∣ x + 1 ∣ : ln ∣ x ∣ - \frac{3}{8} ln 16 x^{2} - 8 x + 8 - \frac{1}{4 7} arctan (\frac{1}{7} (4 x - 1)) - \frac{1}{4} ln ∣ x + 1 ∣

= ln ∣ x ∣ - \frac{3}{8} ln 16 x^{2} - 8 x + 8 - \frac{1}{4 7} arctan (\frac{1}{7} (4 x - 1)) - \frac{1}{4} ln ∣ x + 1 ∣

定数を解答に追加する

= ln ∣ x ∣ - \frac{3}{8} ln 16 x^{2} - 8 x + 8 - \frac{1}{4 7} arctan (\frac{1}{7} (4 x - 1)) - \frac{1}{4} ln ∣ x + 1 ∣ + C