integral of (6x^3-7x+5)/(x^3-x^2)

解

\int \frac{6 x ^{3} - 7 x + 5}{x ^{3} - x ^{2}} d x

6 x + 2 ln ∣ x ∣ + \frac{5}{x} + 4 ln ∣ x - 1 ∣ + C

解答ステップ

\int \frac{6 x ^{3} - 7 x + 5}{x ^{3} - x ^{2}} d x

以下の部分分数を得る:

\frac{6 x ^{3} - 7 x + 5}{x ^{3} - x ^{2}} : 6 + \frac{2}{x} - \frac{5}{x ^{2}} + \frac{4}{x - 1}

= \int 6 + \frac{2}{x} - \frac{5}{x ^{2}} + \frac{4}{x - 1} 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 6 d x + \int \frac{2}{x} d x - \int \frac{5}{x ^{2}} d x + \int \frac{4}{x - 1} d x

\int 6 d x = 6 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

= 6 x + 2 ln ∣ x ∣ - (- \frac{5}{x}) + 4 ln ∣ x - 1 ∣

簡素化

= 6 x + 2 ln ∣ x ∣ + \frac{5}{x} + 4 ln ∣ x - 1 ∣

定数を解答に追加する

= 6 x + 2 ln ∣ x ∣ + \frac{5}{x} + 4 ln ∣ x - 1 ∣ + C