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

解

\int \frac{x + 1}{x ( x + 2 )} d x

\frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C

解答ステップ

\int \frac{x + 1}{x ( x + 2 )} d x

拡張

\frac{x + 1}{x ( x + 2 )} : \frac{x}{x ( x + 2 )} + \frac{1}{x ( x + 2 )}

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 ( x + 2 )} d x + \int \frac{1}{x ( x + 2 )} d x

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

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

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

簡素化

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

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣

定数を解答に追加する

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C