$$$\operatorname{atan}{\left(7 t \right)}$$$ 的積分
您的輸入
求$$$\int \operatorname{atan}{\left(7 t \right)}\, dt$$$。
解答
令 $$$u=7 t$$$。
則 $$$du=\left(7 t\right)^{\prime }dt = 7 dt$$$ (步驟見»),並可得 $$$dt = \frac{du}{7}$$$。
因此,
$${\color{red}{\int{\operatorname{atan}{\left(7 t \right)} d t}}} = {\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{7} d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{7}$$$ 與 $$$f{\left(u \right)} = \operatorname{atan}{\left(u \right)}$$$:
$${\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{7} d u}}} = {\color{red}{\left(\frac{\int{\operatorname{atan}{\left(u \right)} d u}}{7}\right)}}$$
對於積分 $$$\int{\operatorname{atan}{\left(u \right)} d u}$$$,使用分部積分法 $$$\int \operatorname{\omega} \operatorname{dv} = \operatorname{\omega}\operatorname{v} - \int \operatorname{v} \operatorname{d\omega}$$$。
令 $$$\operatorname{\omega}=\operatorname{atan}{\left(u \right)}$$$ 與 $$$\operatorname{dv}=du$$$。
則 $$$\operatorname{d\omega}=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du=\frac{du}{u^{2} + 1}$$$(步驟見 »),且 $$$\operatorname{v}=\int{1 d u}=u$$$(步驟見 »)。
所以,
$$\frac{{\color{red}{\int{\operatorname{atan}{\left(u \right)} d u}}}}{7}=\frac{{\color{red}{\left(\operatorname{atan}{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u^{2} + 1} d u}\right)}}}{7}=\frac{{\color{red}{\left(u \operatorname{atan}{\left(u \right)} - \int{\frac{u}{u^{2} + 1} d u}\right)}}}{7}$$
令 $$$v=u^{2} + 1$$$。
則 $$$dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$$$ (步驟見»),並可得 $$$u du = \frac{dv}{2}$$$。
所以,
$$\frac{u \operatorname{atan}{\left(u \right)}}{7} - \frac{{\color{red}{\int{\frac{u}{u^{2} + 1} d u}}}}{7} = \frac{u \operatorname{atan}{\left(u \right)}}{7} - \frac{{\color{red}{\int{\frac{1}{2 v} d v}}}}{7}$$
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 $$$f{\left(v \right)} = \frac{1}{v}$$$:
$$\frac{u \operatorname{atan}{\left(u \right)}}{7} - \frac{{\color{red}{\int{\frac{1}{2 v} d v}}}}{7} = \frac{u \operatorname{atan}{\left(u \right)}}{7} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{2}\right)}}}{7}$$
$$$\frac{1}{v}$$$ 的積分是 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$\frac{u \operatorname{atan}{\left(u \right)}}{7} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{14} = \frac{u \operatorname{atan}{\left(u \right)}}{7} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{14}$$
回顧一下 $$$v=u^{2} + 1$$$:
$$\frac{u \operatorname{atan}{\left(u \right)}}{7} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{14} = \frac{u \operatorname{atan}{\left(u \right)}}{7} - \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{14}$$
回顧一下 $$$u=7 t$$$:
$$- \frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{14} + \frac{{\color{red}{u}} \operatorname{atan}{\left({\color{red}{u}} \right)}}{7} = - \frac{\ln{\left(1 + {\color{red}{\left(7 t\right)}}^{2} \right)}}{14} + \frac{{\color{red}{\left(7 t\right)}} \operatorname{atan}{\left({\color{red}{\left(7 t\right)}} \right)}}{7}$$
因此,
$$\int{\operatorname{atan}{\left(7 t \right)} d t} = t \operatorname{atan}{\left(7 t \right)} - \frac{\ln{\left(49 t^{2} + 1 \right)}}{14}$$
加上積分常數:
$$\int{\operatorname{atan}{\left(7 t \right)} d t} = t \operatorname{atan}{\left(7 t \right)} - \frac{\ln{\left(49 t^{2} + 1 \right)}}{14}+C$$
答案
$$$\int \operatorname{atan}{\left(7 t \right)}\, dt = \left(t \operatorname{atan}{\left(7 t \right)} - \frac{\ln\left(49 t^{2} + 1\right)}{14}\right) + C$$$A