integral of (8x^2+3x-6)/((x^2-4)(x+2))

解

\int \frac{8 x ^{2} + 3 x - 6}{( x ^{2} - 4 ) ( x + 2 )} d x

6 ln ∣ x + 2 ∣ + \frac{5}{x + 2} + 2 ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{8 x ^{2} + 3 x - 6}{( x ^{2} - 4 ) ( x + 2 )} d x

以下の部分分数を得る:

\frac{8 x ^{2} + 3 x - 6}{( x ^{2} - 4 ) ( x + 2 )} : \frac{6}{x + 2} - \frac{5}{( x + 2 ) ^{2}} + \frac{2}{x - 2}

= \int \frac{6}{x + 2} - \frac{5}{( x + 2 ) ^{2}} + \frac{2}{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

= \int \frac{6}{x + 2} d x - \int \frac{5}{( x + 2 ) ^{2}} d x + \int \frac{2}{x - 2} d x

\int \frac{6}{x + 2} d x = 6 ln ∣ x + 2 ∣

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

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

= 6 ln ∣ x + 2 ∣ - (- \frac{5}{x + 2}) + 2 ln ∣ x - 2 ∣

簡素化

= 6 ln ∣ x + 2 ∣ + \frac{5}{x + 2} + 2 ln ∣ x - 2 ∣

定数を解答に追加する

= 6 ln ∣ x + 2 ∣ + \frac{5}{x + 2} + 2 ln ∣ x - 2 ∣ + C