integral of (6x^2+x-1)/(3x^3-9x^2+x-3)

解

\int \frac{6 x ^{2} + x - 1}{3 x ^{3} - 9 x ^{2} + x - 3} d x

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

解答ステップ

\int \frac{6 x ^{2} + x - 1}{3 x ^{3} - 9 x ^{2} + x - 3} d x

以下の部分分数を得る:

\frac{6 x ^{2} + x - 1}{3 x ^{3} - 9 x ^{2} + x - 3} : \frac{2}{x - 3} + \frac{1}{3 x ^{2} + 1}

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

\int \frac{2}{x - 3} d x = 2 ln ∣ x - 3 ∣

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

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

定数を解答に追加する

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