integral of (2x)/((x+4)(x-5))

解

\int \frac{2 x}{( x + 4 ) ( x - 5 )} d x

2 (\frac{4}{9} ln ∣ x + 4 ∣ + \frac{5}{9} ln ∣ x - 5 ∣) + C

解答ステップ

\int \frac{2 x}{( x + 4 ) ( x - 5 )} d x

定数を除く:

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

= 2 \cdot \int \frac{x}{( x + 4 ) ( x - 5 )} d x

以下の部分分数を得る:

\frac{x}{( x + 4 ) ( x - 5 )} : \frac{4}{9 ( x + 4 )} + \frac{5}{9 ( x - 5 )}

= 2 \cdot \int \frac{4}{9 ( x + 4 )} + \frac{5}{9 ( 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

= 2 (\int \frac{4}{9 ( x + 4 )} d x + \int \frac{5}{9 ( x - 5 )} d x)

\int \frac{4}{9 ( x + 4 )} d x = \frac{4}{9} ln ∣ x + 4 ∣

\int \frac{5}{9 ( x - 5 )} d x = \frac{5}{9} ln ∣ x - 5 ∣

= 2 (\frac{4}{9} ln ∣ x + 4 ∣ + \frac{5}{9} ln ∣ x - 5 ∣)

定数を解答に追加する

= 2 (\frac{4}{9} ln ∣ x + 4 ∣ + \frac{5}{9} ln ∣ x - 5 ∣) + C