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