$$$x^{2} \sec^{2}{\left(x^{3} - 5 \right)}$$$ 的积分

求$$$\int x^{2} \sec^{2}{\left(x^{3} - 5 \right)}\, dx$$$。

设$$$u=x^{3} - 5$$$。

则$$$du=\left(x^{3} - 5\right)^{\prime }dx = 3 x^{2} dx$$$ (步骤见»)，并有$$$x^{2} dx = \frac{du}{3}$$$。

所以，

$${\color{red}{\int{x^{2} \sec^{2}{\left(x^{3} - 5 \right)} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(u \right)}}{3} d u}}}$$

对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(u \right)} = \sec^{2}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{\sec^{2}{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\sec^{2}{\left(u \right)} d u}}{3}\right)}}$$

$$$\sec^{2}{\left(u \right)}$$$ 的积分为 $$$\int{\sec^{2}{\left(u \right)} d u} = \tan{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{\sec^{2}{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\tan{\left(u \right)}}}}{3}$$

回忆一下 $$$u=x^{3} - 5$$$:

$$\frac{\tan{\left({\color{red}{u}} \right)}}{3} = \frac{\tan{\left({\color{red}{\left(x^{3} - 5\right)}} \right)}}{3}$$

因此，

$$\int{x^{2} \sec^{2}{\left(x^{3} - 5 \right)} d x} = \frac{\tan{\left(x^{3} - 5 \right)}}{3}$$

加上积分常数：

$$\int{x^{2} \sec^{2}{\left(x^{3} - 5 \right)} d x} = \frac{\tan{\left(x^{3} - 5 \right)}}{3}+C$$

$$$\int x^{2} \sec^{2}{\left(x^{3} - 5 \right)}\, dx = \frac{\tan{\left(x^{3} - 5 \right)}}{3} + C$$$A