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

解

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

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

解答ステップ

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

拡張

\frac{x ^{4} + x ^{3} - x ^{2} - x + 1}{x ^{3} - x} : \frac{x ^{4}}{x ^{3} - x} + \frac{x ^{3}}{x ^{3} - x} - \frac{x ^{2}}{x ^{3} - x} - \frac{x}{x ^{3} - x} + \frac{1}{x ^{3} - x}

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

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

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

\int \frac{x ^{2}}{x ^{3} - x} d x = \frac{1}{2} ln x^{2} - 1

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

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

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

簡素化

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

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

定数を解答に追加する

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