Integral of $$$p y^{2} \left(p^{2} + 1\right)^{2}$$$ with respect to $$$y$$$
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Your Input
Find $$$\int p y^{2} \left(p^{2} + 1\right)^{2}\, dy$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ with $$$c=p \left(p^{2} + 1\right)^{2}$$$ and $$$f{\left(y \right)} = y^{2}$$$:
$${\color{red}{\int{p y^{2} \left(p^{2} + 1\right)^{2} d y}}} = {\color{red}{p \left(p^{2} + 1\right)^{2} \int{y^{2} d y}}}$$
Apply the power rule $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$p \left(p^{2} + 1\right)^{2} {\color{red}{\int{y^{2} d y}}}=p \left(p^{2} + 1\right)^{2} {\color{red}{\frac{y^{1 + 2}}{1 + 2}}}=p \left(p^{2} + 1\right)^{2} {\color{red}{\left(\frac{y^{3}}{3}\right)}}$$
Therefore,
$$\int{p y^{2} \left(p^{2} + 1\right)^{2} d y} = \frac{p y^{3} \left(p^{2} + 1\right)^{2}}{3}$$
Add the constant of integration:
$$\int{p y^{2} \left(p^{2} + 1\right)^{2} d y} = \frac{p y^{3} \left(p^{2} + 1\right)^{2}}{3}+C$$
Answer
$$$\int p y^{2} \left(p^{2} + 1\right)^{2}\, dy = \frac{p y^{3} \left(p^{2} + 1\right)^{2}}{3} + C$$$A