integral of ((x^2-4x+9))/(x^2-6x+9)

解

\int \frac{( x ^{2} - 4 x + 9 )}{x ^{2} - 6 x + 9} d x

x + 2 ln ∣ x - 3 ∣ - \frac{6}{x - 3} + C

解答ステップ

\int \frac{x ^{2} - 4 x + 9}{x ^{2} - 6 x + 9} d x

拡張

\frac{x ^{2} - 4 x + 9}{x ^{2} - 6 x + 9} : \frac{x ^{2}}{x ^{2} - 6 x + 9} - \frac{4 x}{x ^{2} - 6 x + 9} + \frac{9}{x ^{2} - 6 x + 9}

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 + 9} d x - \int \frac{4 x}{x ^{2} - 6 x + 9} d x + \int \frac{9}{x ^{2} - 6 x + 9} d x

\int \frac{x ^{2}}{x ^{2} - 6 x + 9} d x = 6 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + x

\int \frac{4 x}{x ^{2} - 6 x + 9} d x = 4 (ln ∣ x - 3 ∣ - \frac{3}{x - 3})

\int \frac{9}{x ^{2} - 6 x + 9} d x = - \frac{9}{x - 3}

= 6 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + x - 4 (ln ∣ x - 3 ∣ - \frac{3}{x - 3}) - \frac{9}{x - 3}

簡素化

6 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + x - 4 (ln ∣ x - 3 ∣ - \frac{3}{x - 3}) - \frac{9}{x - 3} : x + 2 ln ∣ x - 3 ∣ - \frac{6}{x - 3}

= x + 2 ln ∣ x - 3 ∣ - \frac{6}{x - 3}

定数を解答に追加する

= x + 2 ln ∣ x - 3 ∣ - \frac{6}{x - 3} + C