$$$\frac{\operatorname{atan}{\left(x \right)}}{4 x^{2}}$$$ 的積分
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您的輸入
求$$$\int \frac{\operatorname{atan}{\left(x \right)}}{4 x^{2}}\, dx$$$。
解答
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(x \right)} = \frac{\operatorname{atan}{\left(x \right)}}{x^{2}}$$$:
$${\color{red}{\int{\frac{\operatorname{atan}{\left(x \right)}}{4 x^{2}} d x}}} = {\color{red}{\left(\frac{\int{\frac{\operatorname{atan}{\left(x \right)}}{x^{2}} d x}}{4}\right)}}$$
令 $$$u=\frac{1}{x}$$$。
則 $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (步驟見»),並可得 $$$\frac{dx}{x^{2}} = - du$$$。
該積分變為
$$\frac{{\color{red}{\int{\frac{\operatorname{atan}{\left(x \right)}}{x^{2}} d x}}}}{4} = \frac{{\color{red}{\int{\left(- \operatorname{acot}{\left(u \right)}\right)d u}}}}{4}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=-1$$$ 與 $$$f{\left(u \right)} = \operatorname{acot}{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\left(- \operatorname{acot}{\left(u \right)}\right)d u}}}}{4} = \frac{{\color{red}{\left(- \int{\operatorname{acot}{\left(u \right)} d u}\right)}}}{4}$$
對於積分 $$$\int{\operatorname{acot}{\left(u \right)} d u}$$$,使用分部積分法 $$$\int \operatorname{\omega} \operatorname{dv} = \operatorname{\omega}\operatorname{v} - \int \operatorname{v} \operatorname{d\omega}$$$。
令 $$$\operatorname{\omega}=\operatorname{acot}{\left(u \right)}$$$ 與 $$$\operatorname{dv}=du$$$。
則 $$$\operatorname{d\omega}=\left(\operatorname{acot}{\left(u \right)}\right)^{\prime }du=- \frac{1}{u^{2} + 1} du$$$(步驟見 »),且 $$$\operatorname{v}=\int{1 d u}=u$$$(步驟見 »)。
該積分變為
$$- \frac{{\color{red}{\int{\operatorname{acot}{\left(u \right)} d u}}}}{4}=- \frac{{\color{red}{\left(\operatorname{acot}{\left(u \right)} \cdot u-\int{u \cdot \left(- \frac{1}{u^{2} + 1}\right) d u}\right)}}}{4}=- \frac{{\color{red}{\left(u \operatorname{acot}{\left(u \right)} - \int{\left(- \frac{u}{u^{2} + 1}\right)d u}\right)}}}{4}$$
令 $$$v=u^{2} + 1$$$。
則 $$$dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$$$ (步驟見»),並可得 $$$u du = \frac{dv}{2}$$$。
該積分變為
$$- \frac{u \operatorname{acot}{\left(u \right)}}{4} + \frac{{\color{red}{\int{\left(- \frac{u}{u^{2} + 1}\right)d u}}}}{4} = - \frac{u \operatorname{acot}{\left(u \right)}}{4} + \frac{{\color{red}{\int{\left(- \frac{1}{2 v}\right)d v}}}}{4}$$
套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$,使用 $$$c=- \frac{1}{2}$$$ 與 $$$f{\left(v \right)} = \frac{1}{v}$$$:
$$- \frac{u \operatorname{acot}{\left(u \right)}}{4} + \frac{{\color{red}{\int{\left(- \frac{1}{2 v}\right)d v}}}}{4} = - \frac{u \operatorname{acot}{\left(u \right)}}{4} + \frac{{\color{red}{\left(- \frac{\int{\frac{1}{v} d v}}{2}\right)}}}{4}$$
$$$\frac{1}{v}$$$ 的積分是 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$- \frac{u \operatorname{acot}{\left(u \right)}}{4} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{8} = - \frac{u \operatorname{acot}{\left(u \right)}}{4} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{8}$$
回顧一下 $$$v=u^{2} + 1$$$:
$$- \frac{u \operatorname{acot}{\left(u \right)}}{4} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{8} = - \frac{u \operatorname{acot}{\left(u \right)}}{4} - \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{8}$$
回顧一下 $$$u=\frac{1}{x}$$$:
$$- \frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{8} - \frac{{\color{red}{u}} \operatorname{acot}{\left({\color{red}{u}} \right)}}{4} = - \frac{\ln{\left(1 + {\color{red}{\frac{1}{x}}}^{2} \right)}}{8} - \frac{{\color{red}{\frac{1}{x}}} \operatorname{acot}{\left({\color{red}{\frac{1}{x}}} \right)}}{4}$$
因此,
$$\int{\frac{\operatorname{atan}{\left(x \right)}}{4 x^{2}} d x} = - \frac{\ln{\left(1 + \frac{1}{x^{2}} \right)}}{8} - \frac{\operatorname{acot}{\left(\frac{1}{x} \right)}}{4 x}$$
化簡:
$$\int{\frac{\operatorname{atan}{\left(x \right)}}{4 x^{2}} d x} = \frac{- x \left(- 2 \ln{\left(x \right)} + \ln{\left(x^{2} + 1 \right)}\right) - 2 \operatorname{atan}{\left(x \right)}}{8 x}$$
加上積分常數:
$$\int{\frac{\operatorname{atan}{\left(x \right)}}{4 x^{2}} d x} = \frac{- x \left(- 2 \ln{\left(x \right)} + \ln{\left(x^{2} + 1 \right)}\right) - 2 \operatorname{atan}{\left(x \right)}}{8 x}+C$$
答案
$$$\int \frac{\operatorname{atan}{\left(x \right)}}{4 x^{2}}\, dx = \frac{- x \left(- 2 \ln\left(x\right) + \ln\left(x^{2} + 1\right)\right) - 2 \operatorname{atan}{\left(x \right)}}{8 x} + C$$$A