$$$x^{2} \operatorname{atan}{\left(4 x \right)}$$$ 的积分

求$$$\int x^{2} \operatorname{atan}{\left(4 x \right)}\, dx$$$。

对于积分$$$\int{x^{2} \operatorname{atan}{\left(4 x \right)} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=\operatorname{atan}{\left(4 x \right)}$$$ 和 $$$\operatorname{dv}=x^{2} dx$$$。

则 $$$\operatorname{du}=\left(\operatorname{atan}{\left(4 x \right)}\right)^{\prime }dx=\frac{4}{16 x^{2} + 1} dx$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{x^{2} d x}=\frac{x^{3}}{3}$$$ (步骤见 »)。

所以，

$${\color{red}{\int{x^{2} \operatorname{atan}{\left(4 x \right)} d x}}}={\color{red}{\left(\operatorname{atan}{\left(4 x \right)} \cdot \frac{x^{3}}{3}-\int{\frac{x^{3}}{3} \cdot \frac{4}{16 x^{2} + 1} d x}\right)}}={\color{red}{\left(\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \int{\frac{4 x^{3}}{48 x^{2} + 3} d x}\right)}}$$

化简被积函数:

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - {\color{red}{\int{\frac{4 x^{3}}{48 x^{2} + 3} d x}}} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - {\color{red}{\int{\frac{4 x^{3}}{3 \left(16 x^{2} + 1\right)} d x}}}$$

对 $$$c=\frac{4}{3}$$$ 和 $$$f{\left(x \right)} = \frac{x^{3}}{16 x^{2} + 1}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - {\color{red}{\int{\frac{4 x^{3}}{3 \left(16 x^{2} + 1\right)} d x}}} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - {\color{red}{\left(\frac{4 \int{\frac{x^{3}}{16 x^{2} + 1} d x}}{3}\right)}}$$

由于分子次数不小于分母次数，进行多项式长除法（步骤见»）:

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{4 {\color{red}{\int{\frac{x^{3}}{16 x^{2} + 1} d x}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{4 {\color{red}{\int{\left(\frac{x}{16} - \frac{x}{16 \left(16 x^{2} + 1\right)}\right)d x}}}}{3}$$

逐项积分：

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{4 {\color{red}{\int{\left(\frac{x}{16} - \frac{x}{16 \left(16 x^{2} + 1\right)}\right)d x}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{4 {\color{red}{\left(\int{\frac{x}{16} d x} - \int{\frac{x}{16 \left(16 x^{2} + 1\right)} d x}\right)}}}{3}$$

对 $$$c=\frac{1}{16}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} + \frac{4 \int{\frac{x}{16 \left(16 x^{2} + 1\right)} d x}}{3} - \frac{4 {\color{red}{\int{\frac{x}{16} d x}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} + \frac{4 \int{\frac{x}{16 \left(16 x^{2} + 1\right)} d x}}{3} - \frac{4 {\color{red}{\left(\frac{\int{x d x}}{16}\right)}}}{3}$$

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

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} + \frac{4 \int{\frac{x}{16 \left(16 x^{2} + 1\right)} d x}}{3} - \frac{{\color{red}{\int{x d x}}}}{12}=\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} + \frac{4 \int{\frac{x}{16 \left(16 x^{2} + 1\right)} d x}}{3} - \frac{{\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{12}=\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} + \frac{4 \int{\frac{x}{16 \left(16 x^{2} + 1\right)} d x}}{3} - \frac{{\color{red}{\left(\frac{x^{2}}{2}\right)}}}{12}$$

设$$$u=256 x^{2} + 16$$$。

则$$$du=\left(256 x^{2} + 16\right)^{\prime }dx = 512 x dx$$$ (步骤见»)，并有$$$x dx = \frac{du}{512}$$$。

该积分可以改写为

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{4 {\color{red}{\int{\frac{x}{16 \left(16 x^{2} + 1\right)} d x}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{4 {\color{red}{\int{\frac{1}{512 u} d u}}}}{3}$$

对 $$$c=\frac{1}{512}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{4 {\color{red}{\int{\frac{1}{512 u} d u}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{4 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{512}\right)}}}{3}$$

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

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{384} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{384}$$

回忆一下 $$$u=256 x^{2} + 16$$$:

$$\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{384} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{\ln{\left(\left|{{\color{red}{\left(256 x^{2} + 16\right)}}}\right| \right)}}{384}$$

因此，

$$\int{x^{2} \operatorname{atan}{\left(4 x \right)} d x} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{\ln{\left(256 x^{2} + 16 \right)}}{384}$$

化简：

$$\int{x^{2} \operatorname{atan}{\left(4 x \right)} d x} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{\ln{\left(16 x^{2} + 1 \right)}}{384} + \frac{\ln{\left(2 \right)}}{96}$$

加上积分常数（并从表达式中去除常数项）：

$$\int{x^{2} \operatorname{atan}{\left(4 x \right)} d x} = \frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{\ln{\left(16 x^{2} + 1 \right)}}{384}+C$$

$$$\int x^{2} \operatorname{atan}{\left(4 x \right)}\, dx = \left(\frac{x^{3} \operatorname{atan}{\left(4 x \right)}}{3} - \frac{x^{2}}{24} + \frac{\ln\left(16 x^{2} + 1\right)}{384}\right) + C$$$A