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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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