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