integral of (4x^2+3x-1)/(x^3-x^2)

解

\int \frac{4 x ^{2} + 3 x - 1}{x ^{3} - x ^{2}} d x

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

解答ステップ

\int \frac{4 x ^{2} + 3 x - 1}{x ^{3} - x ^{2}} d x

拡張

\frac{4 x ^{2} + 3 x - 1}{x ^{3} - x ^{2}} : \frac{4 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{4 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{4 x ^{2}}{x ^{3} - x ^{2}} d x = 4 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 ∣

= 4 ln ∣ x - 1 ∣ + 3 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣)

簡素化

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

= 6 ln ∣ x - 1 ∣ - 2 ln ∣ x ∣ - \frac{1}{x}

定数を解答に追加する

= 6 ln ∣ x - 1 ∣ - 2 ln ∣ x ∣ - \frac{1}{x} + C