$$$\frac{4 x}{x - 6}$$$ 的积分
您的输入
求$$$\int \frac{4 x}{x - 6}\, dx$$$。
解答
对 $$$c=4$$$ 和 $$$f{\left(x \right)} = \frac{x}{x - 6}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$${\color{red}{\int{\frac{4 x}{x - 6} d x}}} = {\color{red}{\left(4 \int{\frac{x}{x - 6} d x}\right)}}$$
改写并拆分该分式:
$$4 {\color{red}{\int{\frac{x}{x - 6} d x}}} = 4 {\color{red}{\int{\left(1 + \frac{6}{x - 6}\right)d x}}}$$
逐项积分:
$$4 {\color{red}{\int{\left(1 + \frac{6}{x - 6}\right)d x}}} = 4 {\color{red}{\left(\int{1 d x} + \int{\frac{6}{x - 6} d x}\right)}}$$
应用常数法则 $$$\int c\, dx = c x$$$,使用 $$$c=1$$$:
$$4 \int{\frac{6}{x - 6} d x} + 4 {\color{red}{\int{1 d x}}} = 4 \int{\frac{6}{x - 6} d x} + 4 {\color{red}{x}}$$
对 $$$c=6$$$ 和 $$$f{\left(x \right)} = \frac{1}{x - 6}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$4 x + 4 {\color{red}{\int{\frac{6}{x - 6} d x}}} = 4 x + 4 {\color{red}{\left(6 \int{\frac{1}{x - 6} d x}\right)}}$$
设$$$u=x - 6$$$。
则$$$du=\left(x - 6\right)^{\prime }dx = 1 dx$$$ (步骤见»),并有$$$dx = du$$$。
该积分可以改写为
$$4 x + 24 {\color{red}{\int{\frac{1}{x - 6} d x}}} = 4 x + 24 {\color{red}{\int{\frac{1}{u} d u}}}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$4 x + 24 {\color{red}{\int{\frac{1}{u} d u}}} = 4 x + 24 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
回忆一下 $$$u=x - 6$$$:
$$4 x + 24 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 4 x + 24 \ln{\left(\left|{{\color{red}{\left(x - 6\right)}}}\right| \right)}$$
因此,
$$\int{\frac{4 x}{x - 6} d x} = 4 x + 24 \ln{\left(\left|{x - 6}\right| \right)}$$
加上积分常数:
$$\int{\frac{4 x}{x - 6} d x} = 4 x + 24 \ln{\left(\left|{x - 6}\right| \right)}+C$$
答案
$$$\int \frac{4 x}{x - 6}\, dx = \left(4 x + 24 \ln\left(\left|{x - 6}\right|\right)\right) + C$$$A