integral of 5v(v^2+2)^2

Lösung

\int 5 v (v^{2} + 2)^{2} d v

5 (\frac{v ^{6}}{6} + v^{4} + 2 v^{2}) + C

Schritte zur Lösung

\int 5 v (v^{2} + 2)^{2} d v

Entferne die Konstante:

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

= 5 \cdot \int v (v^{2} + 2)^{2} d v

Multipliziere aus

v (v^{2} + 2)^{2} : v^{5} + 4 v^{3} + 4 v

= 5 \cdot \int v^{5} + 4 v^{3} + 4 v d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 (\int v^{5} d v + \int 4 v^{3} d v + \int 4 v d v)

\int v^{5} d v = \frac{v ^{6}}{6}

\int 4 v^{3} d v = v^{4}

\int 4 v d v = 2 v^{2}

= 5 (\frac{v ^{6}}{6} + v^{4} + 2 v^{2})

Füge eine Konstante zur Lösung hinzu

= 5 (\frac{v ^{6}}{6} + v^{4} + 2 v^{2}) + C

\frac{d y}{d x} y = \frac{sec ( x )}{1 + sec ( x )}

\int_{0}^{1} \frac{x}{( 2 x + 1 ) ^{3}} d x

\int \frac{1}{x ^{\frac{8}{9}} ( 8 + x ^{\frac{1}{9}} )} d x

\int 4 x^{2} sec (5 x^{3}) d x

\int_{0}^{\frac{π}{4}} 3 tan^{3} (x) d x