Integraal van $$$p y^{2} \left(p^{2} + 1\right)^{2}$$$ met betrekking tot $$$y$$$

Bepaal $$$\int p y^{2} \left(p^{2} + 1\right)^{2}\, dy$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ toe met $$$c=p \left(p^{2} + 1\right)^{2}$$$ en $$$f{\left(y \right)} = y^{2}$$$:

$${\color{red}{\int{p y^{2} \left(p^{2} + 1\right)^{2} d y}}} = {\color{red}{p \left(p^{2} + 1\right)^{2} \int{y^{2} d y}}}$$

Pas de machtsregel $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=2$$$:

$$p \left(p^{2} + 1\right)^{2} {\color{red}{\int{y^{2} d y}}}=p \left(p^{2} + 1\right)^{2} {\color{red}{\frac{y^{1 + 2}}{1 + 2}}}=p \left(p^{2} + 1\right)^{2} {\color{red}{\left(\frac{y^{3}}{3}\right)}}$$

Dus,

$$\int{p y^{2} \left(p^{2} + 1\right)^{2} d y} = \frac{p y^{3} \left(p^{2} + 1\right)^{2}}{3}$$

Voeg de integratieconstante toe:

$$\int{p y^{2} \left(p^{2} + 1\right)^{2} d y} = \frac{p y^{3} \left(p^{2} + 1\right)^{2}}{3}+C$$

$$$\int p y^{2} \left(p^{2} + 1\right)^{2}\, dy = \frac{p y^{3} \left(p^{2} + 1\right)^{2}}{3} + C$$$A