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

解

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

4 (- \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{8} ln ∣ x - 1 ∣) + C

解答ステップ

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

定数を除く:

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

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

以下の部分分数を得る:

\frac{1}{( 2 x + 2 ) ( 2 x - 2 )} : - \frac{1}{8 ( x + 1 )} + \frac{1}{8 ( x - 1 )}

= 4 \cdot \int - \frac{1}{8 ( x + 1 )} + \frac{1}{8 ( x - 1 )} 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

= 4 (- \int \frac{1}{8 ( x + 1 )} d x + \int \frac{1}{8 ( x - 1 )} d x)

\int \frac{1}{8 ( x + 1 )} d x = \frac{1}{8} ln ∣ x + 1 ∣

\int \frac{1}{8 ( x - 1 )} d x = \frac{1}{8} ln ∣ x - 1 ∣

= 4 (- \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{8} ln ∣ x - 1 ∣)

定数を解答に追加する

= 4 (- \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{8} ln ∣ x - 1 ∣) + C