integral of ((x^2-9))/(x-3)

解

\int \frac{( x ^{2} - 9 )}{x - 3} d x

\frac{1}{2} (x - 3)^{2} + 6 x - 18 + C

解答ステップ

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

拡張

\frac{x ^{2} - 9}{x - 3} : \frac{x ^{2}}{x - 3} - \frac{9}{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{x ^{2}}{x - 3} d x - \int \frac{9}{x - 3} d x

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

\int \frac{9}{x - 3} d x = 9 ln ∣ x - 3 ∣

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

簡素化

\frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln ∣ x - 3 ∣ - 9 ln ∣ x - 3 ∣ : \frac{1}{2} (x - 3)^{2} + 6 x - 18

= \frac{1}{2} (x - 3)^{2} + 6 x - 18

定数を解答に追加する

= \frac{1}{2} (x - 3)^{2} + 6 x - 18 + C