integral of ((z^4+10z^3)/(2z^2))

Lösung

\int (\frac{z ^{4} + 10 z ^{3}}{2 z ^{2}}) d z

\frac{1}{2} (\frac{z ^{3}}{3} + 5 z^{2}) + C

Schritte zur Lösung

\int \frac{z ^{4} + 10 z ^{3}}{2 z ^{2}} d z

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int \frac{z ^{4} + 10 z ^{3}}{z ^{2}} d z

\frac{z ^{4} + 10 z ^{3}}{z ^{2}} = z^{2} + 10 z

= \frac{1}{2} \cdot \int z^{2} + 10 z d z

Wende die Summenregel an:

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

= \frac{1}{2} (\int z^{2} d z + \int 10 z d z)

\int z^{2} d z = \frac{z ^{3}}{3}

\int 10 z d z = 5 z^{2}

= \frac{1}{2} (\frac{z ^{3}}{3} + 5 z^{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (\frac{z ^{3}}{3} + 5 z^{2}) + C

y^{^{'}} - \frac{2 y}{x} = x^{2} e^{x}

\frac{\partial}{\partial x} (e^{- k x^{2}})

\frac{d y}{d x} = y (x^{2} + 2)

d er i v a t i v e 8 e^{- 7 x}

y^{^{'}} = x^{3}