integral of (x^{10}-5x^2)^3(x^9-x)

Lösung

\int (x^{10} - 5 x^{2})^{3} (x^{9} - x) d x

\frac{x ^{40}}{40} - \frac{x ^{32}}{2} + \frac{15 x ^{24}}{4} - \frac{25 x ^{16}}{2} + \frac{125 x ^{8}}{8} + C

Schritte zur Lösung

\int (x^{10} - 5 x^{2})^{3} (x^{9} - x) d x

Multipliziere aus

(x^{10} - 5 x^{2})^{3} (x^{9} - x) : x^{39} - 16 x^{31} + 90 x^{23} - 200 x^{15} + 125 x^{7}

= \int x^{39} - 16 x^{31} + 90 x^{23} - 200 x^{15} + 125 x^{7} 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 x^{39} d x - \int 16 x^{31} d x + \int 90 x^{23} d x - \int 200 x^{15} d x + \int 125 x^{7} d x

\int x^{39} d x = \frac{x ^{40}}{40}

\int 16 x^{31} d x = \frac{x ^{32}}{2}

\int 90 x^{23} d x = \frac{15 x ^{24}}{4}

\int 200 x^{15} d x = \frac{25 x ^{16}}{2}

\int 125 x^{7} d x = \frac{125 x ^{8}}{8}

= \frac{x ^{40}}{40} - \frac{x ^{32}}{2} + \frac{15 x ^{24}}{4} - \frac{25 x ^{16}}{2} + \frac{125 x ^{8}}{8}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{40}}{40} - \frac{x ^{32}}{2} + \frac{15 x ^{24}}{4} - \frac{25 x ^{16}}{2} + \frac{125 x ^{8}}{8} + C

x y^{^{'}} + 2 y - 12 = 0

\frac{d}{d x} (ln (\frac{x ^{3}}{e ^{3 x}}))

\int \frac{tan ^{2} ( x )}{x} d x

f (θ) = sec (θ)

(\frac{t}{t + 1 ( t )})^{^{'}}