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

解

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

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

解答ステップ

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

拡張

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

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

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

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

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

簡素化

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

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

定数を解答に追加する

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