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

解

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

ln ∣ x + 1 ∣ - \frac{2}{x + 1} + C

解答ステップ

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

拡張

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

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

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

= ln ∣ x + 1 ∣ + \frac{1}{x + 1} - \frac{3}{x + 1}

簡素化

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

= ln ∣ x + 1 ∣ - \frac{2}{x + 1}

定数を解答に追加する

= ln ∣ x + 1 ∣ - \frac{2}{x + 1} + C