integral of (6x^2-5)/(x-1)

解

\int \frac{6 x ^{2} - 5}{x - 1} d x

3 x^{2} + 6 x + ln ∣ x - 1 ∣ - 9 + C

解答ステップ

\int \frac{6 x ^{2} - 5}{x - 1} d x

拡張

\frac{6 x ^{2} - 5}{x - 1} : \frac{6 x ^{2}}{x - 1} - \frac{5}{x - 1}

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 - 1} d x - \int \frac{5}{x - 1} d x

\int \frac{6 x ^{2}}{x - 1} d x = 6 (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣)

\int \frac{5}{x - 1} d x = 5 ln ∣ x - 1 ∣

= 6 (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣) - 5 ln ∣ x - 1 ∣

簡素化

6 (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣) - 5 ln ∣ x - 1 ∣ : 3 x^{2} + 6 x + ln ∣ x - 1 ∣ - 9

= 3 x^{2} + 6 x + ln ∣ x - 1 ∣ - 9

定数を解答に追加する

= 3 x^{2} + 6 x + ln ∣ x - 1 ∣ - 9 + C