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

Lösung

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

\frac{x ^{6}}{3} + x^{5} + x^{4} + \frac{x ^{3}}{3} + C

Schritte zur Lösung

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

Multipliziere aus

(x^{2} + x)^{2} (2 x + 1) : 2 x^{5} + 5 x^{4} + 4 x^{3} + x^{2}

= \int 2 x^{5} + 5 x^{4} + 4 x^{3} + 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 2 x^{5} d x + \int 5 x^{4} d x + \int 4 x^{3} d x + \int x^{2} d x

\int 2 x^{5} d x = \frac{x ^{6}}{3}

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

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

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

= \frac{x ^{6}}{3} + x^{5} + x^{4} + \frac{x ^{3}}{3}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{6}}{3} + x^{5} + x^{4} + \frac{x ^{3}}{3} + C

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