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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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