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Beliebt Rechnen >

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

Lösung

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

Lösung

2 (\frac{1}{12} ln 2 x + 1 + \frac{1}{12} ln 2 x - 1 + \frac{1}{6} ln x^{2} + 1) + C

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Ermittle den Partialbruch von

\frac{x ^{3}}{2 x ^{4} + x ^{2} - 1} : \frac{1}{6 2 ( 2 x + 1 )} + \frac{1}{6 2 ( 2 x - 1 )} + \frac{x}{3 ( x ^{2} + 1 )}

= 2 \cdot \int \frac{1}{6 2 ( 2 x + 1 )} + \frac{1}{6 2 ( 2 x - 1 )} + \frac{x}{3 ( x ^{2} + 1 )} 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

= 2 (\int \frac{1}{6 2 ( 2 x + 1 )} d x + \int \frac{1}{6 2 ( 2 x - 1 )} d x + \int \frac{x}{3 ( x ^{2} + 1 )} d x)

\int \frac{1}{6 2 ( 2 x + 1 )} d x = \frac{1}{12} ln 2 x + 1

\int \frac{1}{6 2 ( 2 x - 1 )} d x = \frac{1}{12} ln 2 x - 1

\int \frac{x}{3 ( x ^{2} + 1 )} d x = \frac{1}{6} ln x^{2} + 1

= 2 (\frac{1}{12} ln 2 x + 1 + \frac{1}{12} ln 2 x - 1 + \frac{1}{6} ln x^{2} + 1)

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{12} ln 2 x + 1 + \frac{1}{12} ln 2 x - 1 + \frac{1}{6} ln x^{2} + 1) + C

Graph

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