integral of (x-4)/(x^2-3x+2)

解

\int \frac{x - 4}{x ^{2} - 3 x + 2} d x

\frac{1}{2} ln 4 x^{2} - 12 x + 8 + \frac{5}{2} ln ∣ 2 x - 2 ∣ - \frac{5}{2} ln ∣ 2 x - 4 ∣ + C

解答ステップ

\int \frac{x - 4}{x ^{2} - 3 x + 2} d x

拡張

\frac{x - 4}{x ^{2} - 3 x + 2} : \frac{x}{x ^{2} - 3 x + 2} - \frac{4}{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{x}{x ^{2} - 3 x + 2} d x - \int \frac{4}{x ^{2} - 3 x + 2} d x

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

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

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

簡素化

2 (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣)) - (- 8 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)) : \frac{1}{2} ln 4 x^{2} - 12 x + 8 + \frac{5}{2} ln ∣ 2 x - 2 ∣ - \frac{5}{2} ln ∣ 2 x - 4 ∣

= \frac{1}{2} ln 4 x^{2} - 12 x + 8 + \frac{5}{2} ln ∣ 2 x - 2 ∣ - \frac{5}{2} ln ∣ 2 x - 4 ∣

定数を解答に追加する

= \frac{1}{2} ln 4 x^{2} - 12 x + 8 + \frac{5}{2} ln ∣ 2 x - 2 ∣ - \frac{5}{2} ln ∣ 2 x - 4 ∣ + C