$$$y^{23} \left(x + y\right)$$$ 关于$$$x$$$的积分
您的输入
求$$$\int y^{23} \left(x + y\right)\, dx$$$。
解答
对 $$$c=y^{23}$$$ 和 $$$f{\left(x \right)} = x + y$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$${\color{red}{\int{y^{23} \left(x + y\right) d x}}} = {\color{red}{y^{23} \int{\left(x + y\right)d x}}}$$
逐项积分:
$$y^{23} {\color{red}{\int{\left(x + y\right)d x}}} = y^{23} {\color{red}{\left(\int{x d x} + \int{y d x}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=1$$$:
$$y^{23} \left(\int{y d x} + {\color{red}{\int{x d x}}}\right)=y^{23} \left(\int{y d x} + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}\right)=y^{23} \left(\int{y d x} + {\color{red}{\left(\frac{x^{2}}{2}\right)}}\right)$$
应用常数法则 $$$\int c\, dx = c x$$$,使用 $$$c=y$$$:
$$y^{23} \left(\frac{x^{2}}{2} + {\color{red}{\int{y d x}}}\right) = y^{23} \left(\frac{x^{2}}{2} + {\color{red}{x y}}\right)$$
因此,
$$\int{y^{23} \left(x + y\right) d x} = y^{23} \left(\frac{x^{2}}{2} + x y\right)$$
化简:
$$\int{y^{23} \left(x + y\right) d x} = \frac{x y^{23} \left(x + 2 y\right)}{2}$$
加上积分常数:
$$\int{y^{23} \left(x + y\right) d x} = \frac{x y^{23} \left(x + 2 y\right)}{2}+C$$
答案
$$$\int y^{23} \left(x + y\right)\, dx = \frac{x y^{23} \left(x + 2 y\right)}{2} + C$$$A