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

解

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

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

解答ステップ

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

拡張

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

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

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

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

= 6 ln ∣ x - 1 ∣ - 3 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣

簡素化

6 ln ∣ x - 1 ∣ - 3 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣ : 4 ln ∣ x - 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x}

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

定数を解答に追加する

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