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

解

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

- 2 x - \frac{3}{x + 1} + \frac{x ^{2}}{x + 1} - 2 + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

= - 2 x - \frac{3}{x + 1} + \frac{x ^{2}}{x + 1} - 2

定数を解答に追加する

= - 2 x - \frac{3}{x + 1} + \frac{x ^{2}}{x + 1} - 2 + C