$$$\frac{x^{2}}{\left(c - x\right)^{2}}$$$ 关于$$$x$$$的积分

求$$$\int \frac{x^{2}}{\left(c - x\right)^{2}}\, dx$$$。

由于分子次数不小于分母次数，进行多项式长除法:

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

逐项积分：

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

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

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

化简被积函数:

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

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

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

将被积函数的分子改写为 $$$- c + 2 x=-2\left(c - x\right)+c$$$，并将分式拆分:

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

逐项积分：

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

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

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

设$$$u=c - x$$$。

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

因此，

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

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

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

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

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

回忆一下 $$$u=c - x$$$:

$$c \left(2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\frac{c}{\left(c - x\right)^{2}} d x}\right) + x = c \left(2 \ln{\left(\left|{{\color{red}{\left(c - x\right)}}}\right| \right)} + \int{\frac{c}{\left(c - x\right)^{2}} d x}\right) + x$$

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

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

设$$$u=c - x$$$。

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

因此，

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

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

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

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

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

回忆一下 $$$u=c - x$$$:

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

因此，

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

化简：

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

加上积分常数：

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

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