integral of (7x^3-4x+2)/(x^3-x^2)

解

\int \frac{7 x ^{3} - 4 x + 2}{x ^{3} - x ^{2}} d x

\frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{2}{x} + C

解答ステップ

\int \frac{7 x ^{3} - 4 x + 2}{x ^{3} - x ^{2}} d x

拡張

\frac{7 x ^{3} - 4 x + 2}{x ^{3} - x ^{2}} : \frac{7 x ^{3}}{x ^{3} - x ^{2}} - \frac{4 x}{x ^{3} - x ^{2}} + \frac{2}{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{7 x ^{3}}{x ^{3} - x ^{2}} d x - \int \frac{4 x}{x ^{3} - x ^{2}} d x + \int \frac{2}{x ^{3} - x ^{2}} d x

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

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

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

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

簡素化

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

= \frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{2}{x}

定数を解答に追加する

= \frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{2}{x} + C