integral of (7x-1)/((x+2)(x-1)(x-3))

解

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

- ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣ + 2 ln ∣ x - 3 ∣ + C

解答ステップ

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

拡張

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

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

\int \frac{7 x}{( x + 2 ) ( x - 1 ) ( x - 3 )} d x = 7 (- \frac{2}{15} ln ∣ x + 2 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ + \frac{3}{10} ln ∣ x - 3 ∣)

\int \frac{1}{( x + 2 ) ( x - 1 ) ( x - 3 )} d x = \frac{1}{15} ln ∣ x + 2 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{10} ln ∣ x - 3 ∣

= 7 (- \frac{2}{15} ln ∣ x + 2 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ + \frac{3}{10} ln ∣ x - 3 ∣) - (\frac{1}{15} ln ∣ x + 2 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{10} ln ∣ x - 3 ∣)

簡素化

7 (- \frac{2}{15} ln ∣ x + 2 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ + \frac{3}{10} ln ∣ x - 3 ∣) - (\frac{1}{15} ln ∣ x + 2 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{10} ln ∣ x - 3 ∣) : - ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣ + 2 ln ∣ x - 3 ∣

= - ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣ + 2 ln ∣ x - 3 ∣

定数を解答に追加する

= - ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣ + 2 ln ∣ x - 3 ∣ + C