integral of (x^2-6)/((x+2)(2x-1))

解

\int \frac{x ^{2} - 6}{( x + 2 ) ( 2 x - 1 )} d x

\frac{1}{2} x + \frac{2}{5} ln ∣ x + 2 ∣ - \frac{23}{20} ln ∣ 2 x - 1 ∣ + C

解答ステップ

\int \frac{x ^{2} - 6}{( x + 2 ) ( 2 x - 1 )} d x

以下の部分分数を得る:

\frac{x ^{2} - 6}{( x + 2 ) ( 2 x - 1 )} : \frac{1}{2} + \frac{2}{5 ( x + 2 )} - \frac{23}{10 ( 2 x - 1 )}

= \int \frac{1}{2} + \frac{2}{5 ( x + 2 )} - \frac{23}{10 ( 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}{2} d x + \int \frac{2}{5 ( x + 2 )} d x - \int \frac{23}{10 ( 2 x - 1 )} d x

\int \frac{1}{2} d x = \frac{1}{2} x

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

\int \frac{23}{10 ( 2 x - 1 )} d x = \frac{23}{20} ln ∣ 2 x - 1 ∣

= \frac{1}{2} x + \frac{2}{5} ln ∣ x + 2 ∣ - \frac{23}{20} ln ∣ 2 x - 1 ∣

定数を解答に追加する

= \frac{1}{2} x + \frac{2}{5} ln ∣ x + 2 ∣ - \frac{23}{20} ln ∣ 2 x - 1 ∣ + C