integral of 1/((x+1)(x+5))

解

\int \frac{1}{( x + 1 ) ( x + 5 )} d x

\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣ + C

解答ステップ

\int \frac{1}{( x + 1 ) ( x + 5 )} d x

以下の部分分数を得る:

\frac{1}{( x + 1 ) ( x + 5 )} : \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x + 5 )}

= \int \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x + 5 )} 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}{4 ( x + 1 )} d x - \int \frac{1}{4 ( x + 5 )} d x

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

\int \frac{1}{4 ( x + 5 )} d x = \frac{1}{4} ln ∣ x + 5 ∣

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣

定数を解答に追加する

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣ + C