integral of (2x^3+1)^7

Lösung

\int (2 x^{3} + 1)^{7} d x

\frac{64 x ^{22}}{11} + \frac{448 x ^{19}}{19} + 42 x^{16} + \frac{560 x ^{13}}{13} + 28 x^{10} + 12 x^{7} + \frac{7 x ^{4}}{2} + x + C

Schritte zur Lösung

\int (2 x^{3} + 1)^{7} d x

Multipliziere aus

(2 x^{3} + 1)^{7} : 128 x^{21} + 448 x^{18} + 672 x^{15} + 560 x^{12} + 280 x^{9} + 84 x^{6} + 14 x^{3} + 1

= \int 128 x^{21} + 448 x^{18} + 672 x^{15} + 560 x^{12} + 280 x^{9} + 84 x^{6} + 14 x^{3} + 1 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 128 x^{21} d x + \int 448 x^{18} d x + \int 672 x^{15} d x + \int 560 x^{12} d x + \int 280 x^{9} d x + \int 84 x^{6} d x + \int 14 x^{3} d x + \int 1 d x

\int 128 x^{21} d x = \frac{64 x ^{22}}{11}

\int 448 x^{18} d x = \frac{448 x ^{19}}{19}

\int 672 x^{15} d x = 42 x^{16}

\int 560 x^{12} d x = \frac{560 x ^{13}}{13}

\int 280 x^{9} d x = 28 x^{10}

\int 84 x^{6} d x = 12 x^{7}

\int 14 x^{3} d x = \frac{7 x ^{4}}{2}

\int 1 d x = x

= \frac{64 x ^{22}}{11} + \frac{448 x ^{19}}{19} + 42 x^{16} + \frac{560 x ^{13}}{13} + 28 x^{10} + 12 x^{7} + \frac{7 x ^{4}}{2} + x

Füge eine Konstante zur Lösung hinzu

= \frac{64 x ^{22}}{11} + \frac{448 x ^{19}}{19} + 42 x^{16} + \frac{560 x ^{13}}{13} + 28 x^{10} + 12 x^{7} + \frac{7 x ^{4}}{2} + x + C

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