integral of (x^3+3x^2-5x-1)/(x^2-x)

解

\int \frac{x ^{3} + 3 x ^{2} - 5 x - 1}{x ^{2} - x} d x

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

解答ステップ

\int \frac{x ^{3} + 3 x ^{2} - 5 x - 1}{x ^{2} - x} d x

拡張

\frac{x ^{3} + 3 x ^{2} - 5 x - 1}{x ^{2} - x} : \frac{x ^{3}}{x ^{2} - x} + \frac{3 x ^{2}}{x ^{2} - x} - \frac{5 x}{x ^{2} - x} - \frac{1}{x ^{2} - 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 \frac{x ^{3}}{x ^{2} - x} d x + \int \frac{3 x ^{2}}{x ^{2} - x} d x - \int \frac{5 x}{x ^{2} - x} d x - \int \frac{1}{x ^{2} - x} d x

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

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

\int \frac{5 x}{x ^{2} - x} d x = 5 ln ∣ x - 1 ∣

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

= \frac{1}{2} (2 ln (2 ∣ x - 1 ∣) + (x - 1) (x + 3)) + 3 (x - 1 + ln ∣ x - 1 ∣) - 5 ln ∣ x - 1 ∣ - (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

簡素化

\frac{1}{2} (2 ln (2 ∣ x - 1 ∣) + (x - 1) (x + 3)) + 3 (x - 1 + ln ∣ x - 1 ∣) - 5 ln ∣ x - 1 ∣ - (- ln ∣ x ∣ + ln ∣ x - 1 ∣) : \frac{1}{2} x^{2} + 4 x + ln (2 ∣ x - 1 ∣) - 3 ln ∣ x - 1 ∣ + ln ∣ x ∣ - \frac{3}{2} - 3

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

定数を解答に追加する

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