integral of (x^2+2x+3)/(x^2-6x+8)

解

\int \frac{x ^{2} + 2 x + 3}{x ^{2} - 6 x + 8} d x

x - \frac{19}{2} ln ∣ x - 2 ∣ + \frac{19}{2} ln ∣ x - 4 ∣ + 4 ln x^{2} - 6 x + 8 + C

解答ステップ

\int \frac{x ^{2} + 2 x + 3}{x ^{2} - 6 x + 8} d x

拡張

\frac{x ^{2} + 2 x + 3}{x ^{2} - 6 x + 8} : \frac{x ^{2}}{x ^{2} - 6 x + 8} + \frac{2 x}{x ^{2} - 6 x + 8} + \frac{3}{x ^{2} - 6 x + 8}

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 ^{2} - 6 x + 8} d x + \int \frac{2 x}{x ^{2} - 6 x + 8} d x + \int \frac{3}{x ^{2} - 6 x + 8} d x

\int \frac{x ^{2}}{x ^{2} - 6 x + 8} d x = 3 ln x^{2} - 6 x + 8 - 10 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣) + x

\int \frac{2 x}{x ^{2} - 6 x + 8} d x = ln x^{2} - 6 x + 8 - 3 ln ∣ x - 2 ∣ + 3 ln ∣ x - 4 ∣

\int \frac{3}{x ^{2} - 6 x + 8} d x = 3 (- \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 4 ∣)

= 3 ln x^{2} - 6 x + 8 - 10 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣) + x + ln x^{2} - 6 x + 8 - 3 ln ∣ x - 2 ∣ + 3 ln ∣ x - 4 ∣ + 3 (- \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 4 ∣)

簡素化

3 ln x^{2} - 6 x + 8 - 10 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣) + x + ln x^{2} - 6 x + 8 - 3 ln ∣ x - 2 ∣ + 3 ln ∣ x - 4 ∣ + 3 (- \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 4 ∣) : x - \frac{19}{2} ln ∣ x - 2 ∣ + \frac{19}{2} ln ∣ x - 4 ∣ + 4 ln x^{2} - 6 x + 8

= x - \frac{19}{2} ln ∣ x - 2 ∣ + \frac{19}{2} ln ∣ x - 4 ∣ + 4 ln x^{2} - 6 x + 8

定数を解答に追加する

= x - \frac{19}{2} ln ∣ x - 2 ∣ + \frac{19}{2} ln ∣ x - 4 ∣ + 4 ln x^{2} - 6 x + 8 + C