integral of (7-x)/(x^2-4x+4)

解

\int \frac{7 - x}{x ^{2} - 4 x + 4} d x

- \frac{7}{x - 2} + \frac{x}{x - 2} - ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{7 - x}{x ^{2} - 4 x + 4} d x

拡張

\frac{7 - x}{x ^{2} - 4 x + 4} : \frac{7}{x ^{2} - 4 x + 4} - \frac{x}{x ^{2} - 4 x + 4}

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

\int \frac{7}{x ^{2} - 4 x + 4} d x = - \frac{7}{x - 2}

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

= - \frac{7}{x - 2} - (- \frac{x}{x - 2} + ln ∣ x - 2 ∣)

簡素化

= - \frac{7}{x - 2} + \frac{x}{x - 2} - ln ∣ x - 2 ∣

定数を解答に追加する

= - \frac{7}{x - 2} + \frac{x}{x - 2} - ln ∣ x - 2 ∣ + C