integral of (x^2-x+2)/(3x^3-2x^2+6x-4)

解

\int \frac{x ^{2} - x + 2}{3 x ^{3} - 2 x ^{2} + 6 x - 4} d x

\frac{8}{33} ln ∣ 3 x - 2 ∣ + \frac{1}{11} (\frac{1}{2} ln x^{2} + 2 - \frac{3}{2} arctan (\frac{x}{2})) + C

解答ステップ

\int \frac{x ^{2} - x + 2}{3 x ^{3} - 2 x ^{2} + 6 x - 4} d x

以下の部分分数を得る:

\frac{x ^{2} - x + 2}{3 x ^{3} - 2 x ^{2} + 6 x - 4} : \frac{8}{11 ( 3 x - 2 )} + \frac{x - 3}{11 ( x ^{2} + 2 )}

= \int \frac{8}{11 ( 3 x - 2 )} + \frac{x - 3}{11 ( x ^{2} + 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{8}{11 ( 3 x - 2 )} d x + \int \frac{x - 3}{11 ( x ^{2} + 2 )} d x

\int \frac{8}{11 ( 3 x - 2 )} d x = \frac{8}{33} ln ∣ 3 x - 2 ∣

\int \frac{x - 3}{11 ( x ^{2} + 2 )} d x = \frac{1}{11} (\frac{1}{2} ln x^{2} + 2 - \frac{3}{2} arctan (\frac{x}{2}))

= \frac{8}{33} ln ∣ 3 x - 2 ∣ + \frac{1}{11} (\frac{1}{2} ln x^{2} + 2 - \frac{3}{2} arctan (\frac{x}{2}))

定数を解答に追加する

= \frac{8}{33} ln ∣ 3 x - 2 ∣ + \frac{1}{11} (\frac{1}{2} ln x^{2} + 2 - \frac{3}{2} arctan (\frac{x}{2})) + C