integral of ((2x^2-6x+4))/((x-3))

解

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

x^{2} + 4 ln ∣ x - 3 ∣ - 9 + C

解答ステップ

\int \frac{2 x ^{2} - 6 x + 4}{x - 3} d x

拡張

\frac{2 x ^{2} - 6 x + 4}{x - 3} : \frac{2 x ^{2}}{x - 3} - \frac{6 x}{x - 3} + \frac{4}{x - 3}

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{2 x ^{2}}{x - 3} d x - \int \frac{6 x}{x - 3} d x + \int \frac{4}{x - 3} d x

\int \frac{2 x ^{2}}{x - 3} d x = x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27

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

\int \frac{4}{x - 3} d x = 4 ln ∣ x - 3 ∣

= x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27 - 6 (x - 3 + 3 ln ∣ x - 3 ∣) + 4 ln ∣ x - 3 ∣

簡素化

x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27 - 6 (x - 3 + 3 ln ∣ x - 3 ∣) + 4 ln ∣ x - 3 ∣ : x^{2} + 4 ln ∣ x - 3 ∣ - 9

= x^{2} + 4 ln ∣ x - 3 ∣ - 9

定数を解答に追加する

= x^{2} + 4 ln ∣ x - 3 ∣ - 9 + C