integral of 6/(x(x+2))

解

\int \frac{6}{x ( x + 2 )} d x

6 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣) + C

解答ステップ

\int \frac{6}{x ( x + 2 )} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{x ( x + 2 )} d x

以下の部分分数を得る:

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

= 6 \cdot \int \frac{1}{2 x} - \frac{1}{2 ( x + 2 )} d x

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

= 6 (\int \frac{1}{2 x} d x - \int \frac{1}{2 ( x + 2 )} d x)

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

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

= 6 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣)

定数を解答に追加する

= 6 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣) + C