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

解

\int \frac{3 x ^{2} + 2 x - 7}{x ^{2} - 3 x + 2} d x

3 x - 3 ln ∣ x - 1 ∣ + 12 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 + 4 ln ∣ 2 x - 2 ∣ - 4 ln ∣ 2 x - 4 ∣ + C

解答ステップ

\int \frac{3 x ^{2} + 2 x - 7}{x ^{2} - 3 x + 2} d x

拡張

\frac{3 x ^{2} + 2 x - 7}{x ^{2} - 3 x + 2} : \frac{3 x ^{2}}{x ^{2} - 3 x + 2} + \frac{2 x}{x ^{2} - 3 x + 2} - \frac{7}{x ^{2} - 3 x + 2}

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{3 x ^{2}}{x ^{2} - 3 x + 2} d x + \int \frac{2 x}{x ^{2} - 3 x + 2} d x - \int \frac{7}{x ^{2} - 3 x + 2} d x

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

\int \frac{2 x}{x ^{2} - 3 x + 2} d x = ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣

\int \frac{7}{x ^{2} - 3 x + 2} d x = - 14 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)

= 3 (x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣) + ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣ - (- 14 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣))

簡素化

3 (x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣) + ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣ - (- 14 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)) : 3 x - 3 ln ∣ x - 1 ∣ + 12 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 + 4 ln ∣ 2 x - 2 ∣ - 4 ln ∣ 2 x - 4 ∣

= 3 x - 3 ln ∣ x - 1 ∣ + 12 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 + 4 ln ∣ 2 x - 2 ∣ - 4 ln ∣ 2 x - 4 ∣

定数を解答に追加する

= 3 x - 3 ln ∣ x - 1 ∣ + 12 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 + 4 ln ∣ 2 x - 2 ∣ - 4 ln ∣ 2 x - 4 ∣ + C