$$$p y^{2} \left(p^{2} + 1\right)^{2}$$$ 关于$$$y$$$的积分
您的输入
求$$$\int p y^{2} \left(p^{2} + 1\right)^{2}\, dy$$$。
解答
对 $$$c=p \left(p^{2} + 1\right)^{2}$$$ 和 $$$f{\left(y \right)} = y^{2}$$$ 应用常数倍法则 $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$：
$${\color{red}{\int{p y^{2} \left(p^{2} + 1\right)^{2} d y}}} = {\color{red}{p \left(p^{2} + 1\right)^{2} \int{y^{2} d y}}}$$
应用幂法则 $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=2$$$：
$$p \left(p^{2} + 1\right)^{2} {\color{red}{\int{y^{2} d y}}}=p \left(p^{2} + 1\right)^{2} {\color{red}{\frac{y^{1 + 2}}{1 + 2}}}=p \left(p^{2} + 1\right)^{2} {\color{red}{\left(\frac{y^{3}}{3}\right)}}$$
因此，
$$\int{p y^{2} \left(p^{2} + 1\right)^{2} d y} = \frac{p y^{3} \left(p^{2} + 1\right)^{2}}{3}$$
加上积分常数：
$$\int{p y^{2} \left(p^{2} + 1\right)^{2} d y} = \frac{p y^{3} \left(p^{2} + 1\right)^{2}}{3}+C$$
答案
$$$\int p y^{2} \left(p^{2} + 1\right)^{2}\, dy = \frac{p y^{3} \left(p^{2} + 1\right)^{2}}{3} + C$$$A