integral of (6x^2-32x+41)/(x-4)

解

\int \frac{6 x ^{2} - 32 x + 41}{x - 4} d x

3 x^{2} - 8 x + 9 ln ∣ x - 4 ∣ - 16 + C

解答ステップ

\int \frac{6 x ^{2} - 32 x + 41}{x - 4} d x

拡張

\frac{6 x ^{2} - 32 x + 41}{x - 4} : \frac{6 x ^{2}}{x - 4} - \frac{32 x}{x - 4} + \frac{41}{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{6 x ^{2}}{x - 4} d x - \int \frac{32 x}{x - 4} d x + \int \frac{41}{x - 4} d x

\int \frac{6 x ^{2}}{x - 4} d x = 6 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣)

\int \frac{32 x}{x - 4} d x = 32 (x - 4 + 4 ln ∣ x - 4 ∣)

\int \frac{41}{x - 4} d x = 41 ln ∣ x - 4 ∣

= 6 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣) - 32 (x - 4 + 4 ln ∣ x - 4 ∣) + 41 ln ∣ x - 4 ∣

簡素化

6 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣) - 32 (x - 4 + 4 ln ∣ x - 4 ∣) + 41 ln ∣ x - 4 ∣ : 3 x^{2} - 8 x + 9 ln ∣ x - 4 ∣ - 16

= 3 x^{2} - 8 x + 9 ln ∣ x - 4 ∣ - 16

定数を解答に追加する

= 3 x^{2} - 8 x + 9 ln ∣ x - 4 ∣ - 16 + C