integral of (9(x-9))/((x+5)(x-2))

解

\int \frac{9 ( x - 9 )}{( x + 5 ) ( x - 2 )} d x

9 (2 ln ∣ x + 5 ∣ - ln ∣ x - 2 ∣) + C

解答ステップ

\int \frac{9 ( x - 9 )}{( x + 5 ) ( x - 2 )} d x

定数を除く:

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

= 9 \cdot \int \frac{x - 9}{( x + 5 ) ( x - 2 )} d x

以下の部分分数を得る:

\frac{x - 9}{( x + 5 ) ( x - 2 )} : \frac{2}{x + 5} - \frac{1}{x - 2}

= 9 \cdot \int \frac{2}{x + 5} - \frac{1}{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

= 9 (\int \frac{2}{x + 5} d x - \int \frac{1}{x - 2} d x)

\int \frac{2}{x + 5} d x = 2 ln ∣ x + 5 ∣

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

= 9 (2 ln ∣ x + 5 ∣ - ln ∣ x - 2 ∣)

定数を解答に追加する

= 9 (2 ln ∣ x + 5 ∣ - ln ∣ x - 2 ∣) + C