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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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