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

Lösung

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

\frac{8 x ^{7}}{7} + 6 x^{6} + \frac{54 x ^{5}}{5} + \frac{27 x ^{4}}{4} + C

Schritte zur Lösung

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

Multipliziere aus

(2 x^{2} + 3 x)^{3} : 8 x^{6} + 36 x^{5} + 54 x^{4} + 27 x^{3}

= \int 8 x^{6} + 36 x^{5} + 54 x^{4} + 27 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

= \int 8 x^{6} d x + \int 36 x^{5} d x + \int 54 x^{4} d x + \int 27 x^{3} d x

\int 8 x^{6} d x = \frac{8 x ^{7}}{7}

\int 36 x^{5} d x = 6 x^{6}

\int 54 x^{4} d x = \frac{54 x ^{5}}{5}

\int 27 x^{3} d x = \frac{27 x ^{4}}{4}

= \frac{8 x ^{7}}{7} + 6 x^{6} + \frac{54 x ^{5}}{5} + \frac{27 x ^{4}}{4}

Füge eine Konstante zur Lösung hinzu

= \frac{8 x ^{7}}{7} + 6 x^{6} + \frac{54 x ^{5}}{5} + \frac{27 x ^{4}}{4} + C

d er i v a t i v e y = \frac{f ( x )}{x ^{5}}

d er i v a t i v e 2 tan^{2} (7 x)

(2 x y + 3) d x + (x^{2} - 1) d y = 0

3 x y^{^{'}} + 7 y = 0

d er i v a t i v e \frac{e ^{10 x} + e ^{- 10 x}}{x}