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

解

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

6 x + ln ∣ x + 1 ∣ + \frac{1}{3} (x + 1)^{3} - \frac{3}{2} (x + 1)^{2} - \frac{1}{2} (x + 1)^{2} + 6 + C

解答ステップ

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

拡張

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

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

= 6 x + ln ∣ x + 1 ∣ + \frac{1}{3} (x + 1)^{3} - \frac{3}{2} (x + 1)^{2} - \frac{1}{2} (x + 1)^{2} + 6

定数を解答に追加する

= 6 x + ln ∣ x + 1 ∣ + \frac{1}{3} (x + 1)^{3} - \frac{3}{2} (x + 1)^{2} - \frac{1}{2} (x + 1)^{2} + 6 + C