integral of 9x^3(x^3+2x^2+x+8)

Lösung

\int 9 x^{3} (x^{3} + 2 x^{2} + x + 8) d x

9 (\frac{x ^{7}}{7} + \frac{x ^{6}}{3} + \frac{x ^{5}}{5} + 2 x^{4}) + C

Schritte zur Lösung

\int 9 x^{3} (x^{3} + 2 x^{2} + x + 8) d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int x^{3} (x^{3} + 2 x^{2} + x + 8) d x

Multipliziere aus

x^{3} (x^{3} + 2 x^{2} + x + 8) : x^{6} + 2 x^{5} + x^{4} + 8 x^{3}

= 9 \cdot \int x^{6} + 2 x^{5} + x^{4} + 8 x^{3} 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

= 9 (\int x^{6} d x + \int 2 x^{5} d x + \int x^{4} d x + \int 8 x^{3} d x)

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

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

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

\int 8 x^{3} d x = 2 x^{4}

= 9 (\frac{x ^{7}}{7} + \frac{x ^{6}}{3} + \frac{x ^{5}}{5} + 2 x^{4})

Füge eine Konstante zur Lösung hinzu

= 9 (\frac{x ^{7}}{7} + \frac{x ^{6}}{3} + \frac{x ^{5}}{5} + 2 x^{4}) + C

x \to 4 - lim (c x^{2} + 7 x)

\int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x

d er i v a t i v e f (x) = \frac{2}{3 x ^{2}}

\frac{d}{d x} (\frac{4 3 x ^{2}}{x ^{2} - 8 x})

n = 1 \sum \infty 2^{- 2 n}