$$$\sqrt{6} \left(4 x^{7} + 1\right)$$$ 的积分
您的输入
求$$$\int \sqrt{6} \left(4 x^{7} + 1\right)\, dx$$$。
解答
对 $$$c=\sqrt{6}$$$ 和 $$$f{\left(x \right)} = 4 x^{7} + 1$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$${\color{red}{\int{\sqrt{6} \left(4 x^{7} + 1\right) d x}}} = {\color{red}{\sqrt{6} \int{\left(4 x^{7} + 1\right)d x}}}$$
逐项积分：
$$\sqrt{6} {\color{red}{\int{\left(4 x^{7} + 1\right)d x}}} = \sqrt{6} {\color{red}{\left(\int{1 d x} + \int{4 x^{7} d x}\right)}}$$
应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：
$$\sqrt{6} \left(\int{4 x^{7} d x} + {\color{red}{\int{1 d x}}}\right) = \sqrt{6} \left(\int{4 x^{7} d x} + {\color{red}{x}}\right)$$
对 $$$c=4$$$ 和 $$$f{\left(x \right)} = x^{7}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$$\sqrt{6} \left(x + {\color{red}{\int{4 x^{7} d x}}}\right) = \sqrt{6} \left(x + {\color{red}{\left(4 \int{x^{7} d x}\right)}}\right)$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=7$$$：
$$\sqrt{6} \left(x + 4 {\color{red}{\int{x^{7} d x}}}\right)=\sqrt{6} \left(x + 4 {\color{red}{\frac{x^{1 + 7}}{1 + 7}}}\right)=\sqrt{6} \left(x + 4 {\color{red}{\left(\frac{x^{8}}{8}\right)}}\right)$$
因此，
$$\int{\sqrt{6} \left(4 x^{7} + 1\right) d x} = \sqrt{6} \left(\frac{x^{8}}{2} + x\right)$$
化简：
$$\int{\sqrt{6} \left(4 x^{7} + 1\right) d x} = \frac{\sqrt{6} x \left(x^{7} + 2\right)}{2}$$
加上积分常数：
$$\int{\sqrt{6} \left(4 x^{7} + 1\right) d x} = \frac{\sqrt{6} x \left(x^{7} + 2\right)}{2}+C$$
答案
$$$\int \sqrt{6} \left(4 x^{7} + 1\right)\, dx = \frac{\sqrt{6} x \left(x^{7} + 2\right)}{2} + C$$$A