integral of (x^2-3)/((x^2-4)(2x+1))

解

\int \frac{x ^{2} - 3}{( x ^{2} - 4 ) ( 2 x + 1 )} d x

\frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{20} ln ∣ x - 2 ∣ + \frac{11}{30} ln ∣ 2 x + 1 ∣ + C

解答ステップ

\int \frac{x ^{2} - 3}{( x ^{2} - 4 ) ( 2 x + 1 )} d x

以下の部分分数を得る:

\frac{x ^{2} - 3}{( x ^{2} - 4 ) ( 2 x + 1 )} : \frac{1}{12 ( x + 2 )} + \frac{1}{20 ( x - 2 )} + \frac{11}{15 ( 2 x + 1 )}

= \int \frac{1}{12 ( x + 2 )} + \frac{1}{20 ( x - 2 )} + \frac{11}{15 ( 2 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

= \int \frac{1}{12 ( x + 2 )} d x + \int \frac{1}{20 ( x - 2 )} d x + \int \frac{11}{15 ( 2 x + 1 )} d x

\int \frac{1}{12 ( x + 2 )} d x = \frac{1}{12} ln ∣ x + 2 ∣

\int \frac{1}{20 ( x - 2 )} d x = \frac{1}{20} ln ∣ x - 2 ∣

\int \frac{11}{15 ( 2 x + 1 )} d x = \frac{11}{30} ln ∣ 2 x + 1 ∣

= \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{20} ln ∣ x - 2 ∣ + \frac{11}{30} ln ∣ 2 x + 1 ∣

定数を解答に追加する

= \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{20} ln ∣ x - 2 ∣ + \frac{11}{30} ln ∣ 2 x + 1 ∣ + C