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Beliebt Rechnen >

integral of (x^2-8x-4)/(x^3-4x)

Lösung

\int \frac{x ^{2} - 8 x - 4}{x ^{3} - 4 x} d x

Lösung

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

Schritte zur Lösung

\int \frac{x ^{2} - 8 x - 4}{x ^{3} - 4 x} d x

Multipliziere aus

\frac{x ^{2} - 8 x - 4}{x ^{3} - 4 x} : \frac{x ^{2}}{x ^{3} - 4 x} - \frac{8 x}{x ^{3} - 4 x} - \frac{4}{x ^{3} - 4 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 ^{2}}{x ^{3} - 4 x} d x - \int \frac{8 x}{x ^{3} - 4 x} d x - \int \frac{4}{x ^{3} - 4 x} d x

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

\int \frac{8 x}{x ^{3} - 4 x} d x = - 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

Graph

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