integral of (9x^3-6x+4)/(x^3-x^2)

解

\int \frac{9 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} d x

\frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{x} + C

解答ステップ

\int \frac{9 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} d x

拡張

\frac{9 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} : \frac{9 x ^{3}}{x ^{3} - x ^{2}} - \frac{6 x}{x ^{3} - x ^{2}} + \frac{4}{x ^{3} - x ^{2}}

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

\int \frac{9 x ^{3}}{x ^{3} - x ^{2}} d x = 9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1})

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

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

= 9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 4 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣)

簡素化

9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 4 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣) : \frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{x}

= \frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{x}

定数を解答に追加する

= \frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{x} + C