$$$\frac{4 x^{2} - 3}{x^{2}}$$$ 的积分

求$$$\int \frac{4 x^{2} - 3}{x^{2}}\, dx$$$。

Expand the expression:

$${\color{red}{\int{\frac{4 x^{2} - 3}{x^{2}} d x}}} = {\color{red}{\int{\left(4 - \frac{3}{x^{2}}\right)d x}}}$$

逐项积分：

$${\color{red}{\int{\left(4 - \frac{3}{x^{2}}\right)d x}}} = {\color{red}{\left(\int{4 d x} - \int{\frac{3}{x^{2}} d x}\right)}}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=4$$$：

$$- \int{\frac{3}{x^{2}} d x} + {\color{red}{\int{4 d x}}} = - \int{\frac{3}{x^{2}} d x} + {\color{red}{\left(4 x\right)}}$$

对 $$$c=3$$$ 和 $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$4 x - {\color{red}{\int{\frac{3}{x^{2}} d x}}} = 4 x - {\color{red}{\left(3 \int{\frac{1}{x^{2}} d x}\right)}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-2$$$：

$$4 x - 3 {\color{red}{\int{\frac{1}{x^{2}} d x}}}=4 x - 3 {\color{red}{\int{x^{-2} d x}}}=4 x - 3 {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}=4 x - 3 {\color{red}{\left(- x^{-1}\right)}}=4 x - 3 {\color{red}{\left(- \frac{1}{x}\right)}}$$

因此，

$$\int{\frac{4 x^{2} - 3}{x^{2}} d x} = 4 x + \frac{3}{x}$$

加上积分常数：

$$\int{\frac{4 x^{2} - 3}{x^{2}} d x} = 4 x + \frac{3}{x}+C$$

$$$\int \frac{4 x^{2} - 3}{x^{2}}\, dx = \left(4 x + \frac{3}{x}\right) + C$$$A