$$$\frac{1}{11} - \tan^{2}{\left(x \right)}$$$ 的积分

求$$$\int \left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)\, dx$$$。

逐项积分：

$${\color{red}{\int{\left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{11} d x} - \int{\tan^{2}{\left(x \right)} d x}\right)}}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=\frac{1}{11}$$$：

$$- \int{\tan^{2}{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{11} d x}}} = - \int{\tan^{2}{\left(x \right)} d x} + {\color{red}{\left(\frac{x}{11}\right)}}$$

设$$$u=\tan{\left(x \right)}$$$。

则 $$$x=\operatorname{atan}{\left(u \right)}$$$ 且 $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$（步骤见»）。

该积分可以改写为

$$\frac{x}{11} - {\color{red}{\int{\tan^{2}{\left(x \right)} d x}}} = \frac{x}{11} - {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}}$$

改写并拆分该分式:

$$\frac{x}{11} - {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}} = \frac{x}{11} - {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}}$$

逐项积分：

$$\frac{x}{11} - {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}} = \frac{x}{11} - {\color{red}{\left(\int{1 d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}$$

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

$$\frac{x}{11} + \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\int{1 d u}}} = \frac{x}{11} + \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{u}}$$

$$$\frac{1}{u^{2} + 1}$$$ 的积分为 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$- u + \frac{x}{11} + {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = - u + \frac{x}{11} + {\color{red}{\operatorname{atan}{\left(u \right)}}}$$

回忆一下 $$$u=\tan{\left(x \right)}$$$:

$$\frac{x}{11} + \operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}} = \frac{x}{11} + \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} - {\color{red}{\tan{\left(x \right)}}}$$

因此，

$$\int{\left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)d x} = \frac{x}{11} - \tan{\left(x \right)} + \operatorname{atan}{\left(\tan{\left(x \right)} \right)}$$

化简：

$$\int{\left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)d x} = \frac{12 x}{11} - \tan{\left(x \right)}$$

加上积分常数：

$$\int{\left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)d x} = \frac{12 x}{11} - \tan{\left(x \right)}+C$$

$$$\int \left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)\, dx = \left(\frac{12 x}{11} - \tan{\left(x \right)}\right) + C$$$A