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

Lösung

\int (4 x + 3) (2 x^{2} + 3 x)^{2} d x

\frac{8 x ^{6}}{3} + 12 x^{5} + 18 x^{4} + 9 x^{3} + C

Schritte zur Lösung

\int (4 x + 3) (2 x^{2} + 3 x)^{2} d x

Multipliziere aus

(4 x + 3) (2 x^{2} + 3 x)^{2} : 16 x^{5} + 60 x^{4} + 72 x^{3} + 27 x^{2}

= \int 16 x^{5} + 60 x^{4} + 72 x^{3} + 27 x^{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 16 x^{5} d x + \int 60 x^{4} d x + \int 72 x^{3} d x + \int 27 x^{2} d x

\int 16 x^{5} d x = \frac{8 x ^{6}}{3}

\int 60 x^{4} d x = 12 x^{5}

\int 72 x^{3} d x = 18 x^{4}

\int 27 x^{2} d x = 9 x^{3}

= \frac{8 x ^{6}}{3} + 12 x^{5} + 18 x^{4} + 9 x^{3}

Füge eine Konstante zur Lösung hinzu

= \frac{8 x ^{6}}{3} + 12 x^{5} + 18 x^{4} + 9 x^{3} + C

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