integral of (8x^2+4x+9)/(x^3-1)

解

\int \frac{8 x ^{2} + 4 x + 9}{x ^{3} - 1} d x

7 ln ∣ x - 1 ∣ + \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1)) + C

解答ステップ

\int \frac{8 x ^{2} + 4 x + 9}{x ^{3} - 1} d x

以下の部分分数を得る:

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

= \int \frac{7}{x - 1} + \frac{x - 2}{x ^{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{7}{x - 1} d x + \int \frac{x - 2}{x ^{2} + x + 1} d x

\int \frac{7}{x - 1} d x = 7 ln ∣ x - 1 ∣

\int \frac{x - 2}{x ^{2} + x + 1} d x = \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1))

= 7 ln ∣ x - 1 ∣ + \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1))

定数を解答に追加する

= 7 ln ∣ x - 1 ∣ + \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1)) + C