integral of (x^2-4)^4(2x-1)

Lösung

\int (x^{2} - 4)^{4} (2 x - 1) d x

\frac{x ^{10}}{5} - \frac{x ^{9}}{9} - 4 x^{8} + \frac{16 x ^{7}}{7} + 32 x^{6} - \frac{96 x ^{5}}{5} - 128 x^{4} + \frac{256 x ^{3}}{3} + 256 x^{2} - 256 x + C

Schritte zur Lösung

\int (x^{2} - 4)^{4} (2 x - 1) d x

Multipliziere aus

(x^{2} - 4)^{4} (2 x - 1) : 2 x^{9} - x^{8} - 32 x^{7} + 16 x^{6} + 192 x^{5} - 96 x^{4} - 512 x^{3} + 256 x^{2} + 512 x - 256

= \int 2 x^{9} - x^{8} - 32 x^{7} + 16 x^{6} + 192 x^{5} - 96 x^{4} - 512 x^{3} + 256 x^{2} + 512 x - 256 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^{9} d x - \int x^{8} d x - \int 32 x^{7} d x + \int 16 x^{6} d x + \int 192 x^{5} d x - \int 96 x^{4} d x - \int 512 x^{3} d x + \int 256 x^{2} d x + \int 512 x d x - \int 256 d x

\int 2 x^{9} d x = \frac{x ^{10}}{5}

\int x^{8} d x = \frac{x ^{9}}{9}

\int 32 x^{7} d x = 4 x^{8}

\int 16 x^{6} d x = \frac{16 x ^{7}}{7}

\int 192 x^{5} d x = 32 x^{6}

\int 96 x^{4} d x = \frac{96 x ^{5}}{5}

\int 512 x^{3} d x = 128 x^{4}

\int 256 x^{2} d x = \frac{256 x ^{3}}{3}

\int 512 x d x = 256 x^{2}

\int 256 d x = 256 x

= \frac{x ^{10}}{5} - \frac{x ^{9}}{9} - 4 x^{8} + \frac{16 x ^{7}}{7} + 32 x^{6} - \frac{96 x ^{5}}{5} - 128 x^{4} + \frac{256 x ^{3}}{3} + 256 x^{2} - 256 x

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{10}}{5} - \frac{x ^{9}}{9} - 4 x^{8} + \frac{16 x ^{7}}{7} + 32 x^{6} - \frac{96 x ^{5}}{5} - 128 x^{4} + \frac{256 x ^{3}}{3} + 256 x^{2} - 256 x + C

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