integral of 1/(x^4+2x^2-8)

Lösung

\int \frac{1}{x ^{4} + 2 x ^{2} - 8} d x

- \frac{1}{12 2} ln x + 2 + \frac{1}{12 2} ln x - 2 - \frac{1}{12} arctan (\frac{x}{2}) + C

Schritte zur Lösung

\int \frac{1}{x ^{4} + 2 x ^{2} - 8} d x

Ermittle den Partialbruch von

\frac{1}{x ^{4} + 2 x ^{2} - 8} : - \frac{1}{12 2 ( x + 2 )} + \frac{1}{12 2 ( x - 2 )} - \frac{1}{6 ( x ^{2} + 4 )}

= \int - \frac{1}{12 2 ( x + 2 )} + \frac{1}{12 2 ( x - 2 )} - \frac{1}{6 ( x ^{2} + 4 )} d 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{1}{12 2 ( x + 2 )} d x + \int \frac{1}{12 2 ( x - 2 )} d x - \int \frac{1}{6 ( x ^{2} + 4 )} d x

\int \frac{1}{12 2 ( x + 2 )} d x = \frac{1}{12 2} ln x + 2

\int \frac{1}{12 2 ( x - 2 )} d x = \frac{1}{12 2} ln x - 2

\int \frac{1}{6 ( x ^{2} + 4 )} d x = \frac{1}{12} arctan (\frac{x}{2})

= - \frac{1}{12 2} ln x + 2 + \frac{1}{12 2} ln x - 2 - \frac{1}{12} arctan (\frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{12 2} ln x + 2 + \frac{1}{12 2} ln x - 2 - \frac{1}{12} arctan (\frac{x}{2}) + C

\int - e^{- u} d u

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