integral of ((3x^4-2x^2+2))/(x^3(x-1))

解

\int \frac{( 3 x ^{4} - 2 x ^{2} + 2 )}{x ^{3} ( x - 1 )} d x

3 x + 3 ln ∣ x - 1 ∣ + \frac{2}{x} + \frac{1}{x ^{2}} - 3 + C

解答ステップ

\int \frac{3 x ^{4} - 2 x ^{2} + 2}{x ^{3} ( x - 1 )} d x

拡張

\frac{3 x ^{4} - 2 x ^{2} + 2}{x ^{3} ( x - 1 )} : \frac{3 x ^{4}}{x ^{3} ( x - 1 )} - \frac{2 x ^{2}}{x ^{3} ( x - 1 )} + \frac{2}{x ^{3} ( 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{3 x ^{4}}{x ^{3} ( x - 1 )} d x - \int \frac{2 x ^{2}}{x ^{3} ( x - 1 )} d x + \int \frac{2}{x ^{3} ( x - 1 )} d x

\int \frac{3 x ^{4}}{x ^{3} ( x - 1 )} d x = 3 (x - 1 + ln ∣ x - 1 ∣)

\int \frac{2 x ^{2}}{x ^{3} ( x - 1 )} d x = 2 (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

\int \frac{2}{x ^{3} ( x - 1 )} d x = - 2 ln ∣ x ∣ + \frac{2}{x} + \frac{1}{x ^{2}} + 2 ln ∣ x - 1 ∣

= 3 (x - 1 + ln ∣ x - 1 ∣) - 2 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - 2 ln ∣ x ∣ + \frac{2}{x} + \frac{1}{x ^{2}} + 2 ln ∣ x - 1 ∣

簡素化

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

= 3 x + 3 ln ∣ x - 1 ∣ + \frac{2}{x} + \frac{1}{x ^{2}} - 3

定数を解答に追加する

= 3 x + 3 ln ∣ x - 1 ∣ + \frac{2}{x} + \frac{1}{x ^{2}} - 3 + C