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

解

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

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

解答ステップ

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

拡張

\frac{u + 1}{u ( u + 2 )} : \frac{u}{u ( u + 2 )} + \frac{1}{u ( u + 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{u}{u ( u + 2 )} d u + \int \frac{1}{u ( u + 2 )} d u

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

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

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

簡素化

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

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

定数を解答に追加する

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