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

Lösung

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

\frac{3 x ^{2}}{2} + \frac{3 x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{2 x ^{5}}{5} + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int 3 x d x = \frac{3 x ^{2}}{2}

\int 3 x^{3} d x = \frac{3 x ^{4}}{4}

\int 2 x^{2} d x = \frac{2 x ^{3}}{3}

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

= \frac{3 x ^{2}}{2} + \frac{3 x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{2 x ^{5}}{5}

Füge eine Konstante zur Lösung hinzu

= \frac{3 x ^{2}}{2} + \frac{3 x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{2 x ^{5}}{5} + C

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