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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

\int k x x d x

\int (x^{2} - 1)^{9} x d x

\int 100 (0.6)^{x} d x

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\int 13 x^{2} arctan (x) d x