$$$\frac{x}{2 x^{2} - 1}$$$ 的积分

求$$$\int \frac{x}{2 x^{2} - 1}\, dx$$$。

设$$$u=2 x^{2} - 1$$$。

则$$$du=\left(2 x^{2} - 1\right)^{\prime }dx = 4 x dx$$$ (步骤见»)，并有$$$x dx = \frac{du}{4}$$$。

该积分可以改写为

$${\color{red}{\int{\frac{x}{2 x^{2} - 1} d x}}} = {\color{red}{\int{\frac{1}{4 u} d u}}}$$

对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{1}{4 u} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{4}\right)}}$$

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

回忆一下 $$$u=2 x^{2} - 1$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} = \frac{\ln{\left(\left|{{\color{red}{\left(2 x^{2} - 1\right)}}}\right| \right)}}{4}$$

因此，

$$\int{\frac{x}{2 x^{2} - 1} d x} = \frac{\ln{\left(\left|{2 x^{2} - 1}\right| \right)}}{4}$$

加上积分常数：

$$\int{\frac{x}{2 x^{2} - 1} d x} = \frac{\ln{\left(\left|{2 x^{2} - 1}\right| \right)}}{4}+C$$

$$$\int \frac{x}{2 x^{2} - 1}\, dx = \frac{\ln\left(\left|{2 x^{2} - 1}\right|\right)}{4} + C$$$A