integral of 1/((x-2)(x+4))

解

\int \frac{1}{( x - 2 ) ( x + 4 )} d x

\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣ + C

解答ステップ

\int \frac{1}{( x - 2 ) ( x + 4 )} d x

以下の部分分数を得る:

\frac{1}{( x - 2 ) ( x + 4 )} : \frac{1}{6 ( x - 2 )} - \frac{1}{6 ( x + 4 )}

= \int \frac{1}{6 ( x - 2 )} - \frac{1}{6 ( x + 4 )} 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

= \int \frac{1}{6 ( x - 2 )} d x - \int \frac{1}{6 ( x + 4 )} d x

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

\int \frac{1}{6 ( x + 4 )} d x = \frac{1}{6} ln ∣ x + 4 ∣

= \frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣

定数を解答に追加する

= \frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣ + C