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$$$\frac{1}{9 x^{2} - 4}$$$ 的积分
您的输入
求$$$\int \frac{1}{9 x^{2} - 4}\, dx$$$。
解答
进行部分分式分解（步骤可见»）:
$${\color{red}{\int{\frac{1}{9 x^{2} - 4} d x}}} = {\color{red}{\int{\left(- \frac{1}{4 \left(3 x + 2\right)} + \frac{1}{4 \left(3 x - 2\right)}\right)d x}}}$$
逐项积分：
$${\color{red}{\int{\left(- \frac{1}{4 \left(3 x + 2\right)} + \frac{1}{4 \left(3 x - 2\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{4 \left(3 x - 2\right)} d x} - \int{\frac{1}{4 \left(3 x + 2\right)} d x}\right)}}$$
对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(x \right)} = \frac{1}{3 x + 2}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$$\int{\frac{1}{4 \left(3 x - 2\right)} d x} - {\color{red}{\int{\frac{1}{4 \left(3 x + 2\right)} d x}}} = \int{\frac{1}{4 \left(3 x - 2\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{3 x + 2} d x}}{4}\right)}}$$
设$$$u=3 x + 2$$$。
则$$$du=\left(3 x + 2\right)^{\prime }dx = 3 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{3}$$$。
该积分可以改写为
$$\int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{3 x + 2} d x}}}}{4} = \int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4}$$
对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：
$$\int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4} = \int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{4}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{12} = \int{\frac{1}{4 \left(3 x - 2\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{12}$$
回忆一下 $$$u=3 x + 2$$$:
$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{12} + \int{\frac{1}{4 \left(3 x - 2\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(3 x + 2\right)}}}\right| \right)}}{12} + \int{\frac{1}{4 \left(3 x - 2\right)} d x}$$
对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(x \right)} = \frac{1}{3 x - 2}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + {\color{red}{\int{\frac{1}{4 \left(3 x - 2\right)} d x}}} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + {\color{red}{\left(\frac{\int{\frac{1}{3 x - 2} d x}}{4}\right)}}$$
设$$$u=3 x - 2$$$。
则$$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{3}$$$。
因此，
$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{3 x - 2} d x}}}}{4} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4}$$
对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：
$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{4}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{12} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{12}$$
回忆一下 $$$u=3 x - 2$$$:
$$- \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{12} = - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12} + \frac{\ln{\left(\left|{{\color{red}{\left(3 x - 2\right)}}}\right| \right)}}{12}$$
因此，
$$\int{\frac{1}{9 x^{2} - 4} d x} = \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12}$$
加上积分常数：
$$\int{\frac{1}{9 x^{2} - 4} d x} = \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} - \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12}+C$$
答案
$$$\int \frac{1}{9 x^{2} - 4}\, dx = \left(\frac{\ln\left(\left|{3 x - 2}\right|\right)}{12} - \frac{\ln\left(\left|{3 x + 2}\right|\right)}{12}\right) + C$$$A