integral of (2x^4-18)/(x^2+3)

Lösung

\int \frac{2 x ^{4} - 18}{x ^{2} + 3} d x

\frac{2 x ^{3}}{3} - 6 x + C

Schritte zur Lösung

\int \frac{2 x ^{4} - 18}{x ^{2} + 3} d x

Multipliziere aus

\frac{2 x ^{4} - 18}{x ^{2} + 3} : \frac{2 x ^{4}}{x ^{2} + 3} - \frac{18}{x ^{2} + 3}

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{2 x ^{4}}{x ^{2} + 3} d x - \int \frac{18}{x ^{2} + 3} d x

\int \frac{2 x ^{4}}{x ^{2} + 3} d x = \frac{2 x ^{3}}{3} - 6 x + 63 arctan (\frac{x}{3})

\int \frac{18}{x ^{2} + 3} d x = 63 arctan (\frac{x}{3})

= \frac{2 x ^{3}}{3} - 6 x + 63 arctan (\frac{x}{3}) - 63 arctan (\frac{x}{3})

Vereinfache

= \frac{2 x ^{3}}{3} - 6 x

Füge eine Konstante zur Lösung hinzu

= \frac{2 x ^{3}}{3} - 6 x + C

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