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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{x ^{5} - x} d x - \int \frac{2}{x ^{5} - x} d x

\int \frac{x}{x ^{5} - x} d x = - \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{2}{x ^{5} - x} d x = 2 (- ln ∣ x ∣ + \frac{1}{4} ln x^{2} + 1 + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣)

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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

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