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

解

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

\frac{5}{2} ln ∣ x - 1 ∣ + \frac{19}{2} ln ∣ x - 3 ∣ - 11 ln ∣ x - 2 ∣ + C

解答ステップ

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

拡張

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

\int \frac{x ^{2}}{( x - 1 ) ( x - 2 ) ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 1 ∣ - 4 ln ∣ x - 2 ∣ + \frac{9}{2} ln ∣ x - 3 ∣

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

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

= \frac{1}{2} ln ∣ x - 1 ∣ - 4 ln ∣ x - 2 ∣ + \frac{9}{2} ln ∣ x - 3 ∣ + 3 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{3}{2} ln ∣ x - 3 ∣ - 2 ln ∣ x - 2 ∣) + \frac{1}{2} ln ∣ x - 1 ∣ - ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 3 ∣

簡素化

\frac{1}{2} ln ∣ x - 1 ∣ - 4 ln ∣ x - 2 ∣ + \frac{9}{2} ln ∣ x - 3 ∣ + 3 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{3}{2} ln ∣ x - 3 ∣ - 2 ln ∣ x - 2 ∣) + \frac{1}{2} ln ∣ x - 1 ∣ - ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ : \frac{5}{2} ln ∣ x - 1 ∣ + \frac{19}{2} ln ∣ x - 3 ∣ - 11 ln ∣ x - 2 ∣

= \frac{5}{2} ln ∣ x - 1 ∣ + \frac{19}{2} ln ∣ x - 3 ∣ - 11 ln ∣ x - 2 ∣

定数を解答に追加する

= \frac{5}{2} ln ∣ x - 1 ∣ + \frac{19}{2} ln ∣ x - 3 ∣ - 11 ln ∣ x - 2 ∣ + C