integral of (x+4)/(x^2+4x+4)

解

\int \frac{x + 4}{x ^{2} + 4 x + 4} d x

ln ∣ x + 2 ∣ - \frac{2}{x + 2} + C

解答ステップ

\int \frac{x + 4}{x ^{2} + 4 x + 4} d x

拡張

\frac{x + 4}{x ^{2} + 4 x + 4} : \frac{x}{x ^{2} + 4 x + 4} + \frac{4}{x ^{2} + 4 x + 4}

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} + 4 x + 4} d x + \int \frac{4}{x ^{2} + 4 x + 4} d x

\int \frac{x}{x ^{2} + 4 x + 4} d x = ln ∣ x + 2 ∣ + \frac{2}{x + 2}

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

= ln ∣ x + 2 ∣ + \frac{2}{x + 2} - \frac{4}{x + 2}

簡素化

ln ∣ x + 2 ∣ + \frac{2}{x + 2} - \frac{4}{x + 2} : ln ∣ x + 2 ∣ - \frac{2}{x + 2}

= ln ∣ x + 2 ∣ - \frac{2}{x + 2}

定数を解答に追加する

= ln ∣ x + 2 ∣ - \frac{2}{x + 2} + C