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

解

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

\frac{2}{7} ln ∣ x + 2 ∣ + \frac{1}{21} ln ∣ 3 x - 1 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{x}{( x + 2 ) ( 3 x - 1 )} : \frac{2}{7 ( x + 2 )} + \frac{1}{7 ( 3 x - 1 )}

= \int \frac{2}{7 ( x + 2 )} + \frac{1}{7 ( 3 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{2}{7 ( x + 2 )} d x + \int \frac{1}{7 ( 3 x - 1 )} d x

\int \frac{2}{7 ( x + 2 )} d x = \frac{2}{7} ln ∣ x + 2 ∣

\int \frac{1}{7 ( 3 x - 1 )} d x = \frac{1}{21} ln ∣ 3 x - 1 ∣

= \frac{2}{7} ln ∣ x + 2 ∣ + \frac{1}{21} ln ∣ 3 x - 1 ∣

定数を解答に追加する

= \frac{2}{7} ln ∣ x + 2 ∣ + \frac{1}{21} ln ∣ 3 x - 1 ∣ + C