integral of p(p+1)^4

Lösung

\int p (p + 1)^{4} d p

\frac{p ^{6}}{6} + \frac{4 p ^{5}}{5} + \frac{3 p ^{4}}{2} + \frac{4 p ^{3}}{3} + \frac{p ^{2}}{2} + C

Schritte zur Lösung

\int p (p + 1)^{4} d p

Multipliziere aus

p (p + 1)^{4} : p^{5} + 4 p^{4} + 6 p^{3} + 4 p^{2} + p

= \int p^{5} + 4 p^{4} + 6 p^{3} + 4 p^{2} + p d p

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int p^{5} d p + \int 4 p^{4} d p + \int 6 p^{3} d p + \int 4 p^{2} d p + \int p d p

\int p^{5} d p = \frac{p ^{6}}{6}

\int 4 p^{4} d p = \frac{4 p ^{5}}{5}

\int 6 p^{3} d p = \frac{3 p ^{4}}{2}

\int 4 p^{2} d p = \frac{4 p ^{3}}{3}

\int p d p = \frac{p ^{2}}{2}

= \frac{p ^{6}}{6} + \frac{4 p ^{5}}{5} + \frac{3 p ^{4}}{2} + \frac{4 p ^{3}}{3} + \frac{p ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= \frac{p ^{6}}{6} + \frac{4 p ^{5}}{5} + \frac{3 p ^{4}}{2} + \frac{4 p ^{3}}{3} + \frac{p ^{2}}{2} + C

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