$$$\sec^{2}{\left(x + 1 \right)}$$$ 的積分

求$$$\int \sec^{2}{\left(x + 1 \right)}\, dx$$$。

令 $$$u=x + 1$$$。

則 $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

該積分變為

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

$$$\sec^{2}{\left(u \right)}$$$ 的積分是 $$$\int{\sec^{2}{\left(u \right)} d u} = \tan{\left(u \right)}$$$：

$${\color{red}{\int{\sec^{2}{\left(u \right)} d u}}} = {\color{red}{\tan{\left(u \right)}}}$$

回顧一下 $$$u=x + 1$$$：

$$\tan{\left({\color{red}{u}} \right)} = \tan{\left({\color{red}{\left(x + 1\right)}} \right)}$$

因此，

$$\int{\sec^{2}{\left(x + 1 \right)} d x} = \tan{\left(x + 1 \right)}$$

加上積分常數：

$$\int{\sec^{2}{\left(x + 1 \right)} d x} = \tan{\left(x + 1 \right)}+C$$

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