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

解

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

\frac{1}{7} (- ln 4 x^{2} - 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x - 1)) + 23 arctan (\frac{1}{3} (2 x - 1))) + \frac{2}{7} ln ∣ 3 x - 2 ∣ + C

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{- 2 x + 3}{7 ( x ^{2} - x + 1 )} d x = \frac{1}{7} (- ln 4 x^{2} - 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x - 1)) + 23 arctan (\frac{1}{3} (2 x - 1)))

\int \frac{6}{7 ( 3 x - 2 )} d x = \frac{2}{7} ln ∣ 3 x - 2 ∣

= \frac{1}{7} (- ln 4 x^{2} - 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x - 1)) + 23 arctan (\frac{1}{3} (2 x - 1))) + \frac{2}{7} ln ∣ 3 x - 2 ∣

定数を解答に追加する

= \frac{1}{7} (- ln 4 x^{2} - 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x - 1)) + 23 arctan (\frac{1}{3} (2 x - 1))) + \frac{2}{7} ln ∣ 3 x - 2 ∣ + C