integral of (4x^2-7x-12)/(x(x+2)(x-3))

解

\int \frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} d x

2 ln ∣ x ∣ + \frac{9}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣ + C

解答ステップ

\int \frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} d x

以下の部分分数を得る:

\frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} : \frac{2}{x} + \frac{9}{5 ( x + 2 )} + \frac{1}{5 ( x - 3 )}

= \int \frac{2}{x} + \frac{9}{5 ( x + 2 )} + \frac{1}{5 ( x - 3 )} 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{2}{x} d x + \int \frac{9}{5 ( x + 2 )} d x + \int \frac{1}{5 ( x - 3 )} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{9}{5 ( x + 2 )} d x = \frac{9}{5} ln ∣ x + 2 ∣

\int \frac{1}{5 ( x - 3 )} d x = \frac{1}{5} ln ∣ x - 3 ∣

= 2 ln ∣ x ∣ + \frac{9}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣

定数を解答に追加する

= 2 ln ∣ x ∣ + \frac{9}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣ + C