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

解

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

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

解答ステップ

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

拡張

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

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

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

簡素化

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

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

定数を解答に追加する

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