$$$\frac{\sqrt{57} int_{0}^{4} x^{2} \sqrt{x^{3}}}{4}$$$ 关于$$$x$$$的积分

求$$$\int \frac{\sqrt{57} int_{0}^{4} x^{2} \sqrt{x^{3}}}{4}\, dx$$$。

输入已重写为：$$$\int{\frac{\sqrt{57} int_{0}^{4} x^{2} \sqrt{x^{3}}}{4} d x}=\int{\frac{\sqrt{57} int_{0}^{4} x^{\frac{7}{2}}}{4} d x}$$$。

对 $$$c=\frac{\sqrt{57} int_{0}^{4}}{4}$$$ 和 $$$f{\left(x \right)} = x^{\frac{7}{2}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{\sqrt{57} int_{0}^{4} x^{\frac{7}{2}}}{4} d x}}} = {\color{red}{\left(\frac{\sqrt{57} int_{0}^{4} \int{x^{\frac{7}{2}} d x}}{4}\right)}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=\frac{7}{2}$$$：

$$\frac{\sqrt{57} int_{0}^{4} {\color{red}{\int{x^{\frac{7}{2}} d x}}}}{4}=\frac{\sqrt{57} int_{0}^{4} {\color{red}{\frac{x^{1 + \frac{7}{2}}}{1 + \frac{7}{2}}}}}{4}=\frac{\sqrt{57} int_{0}^{4} {\color{red}{\left(\frac{2 x^{\frac{9}{2}}}{9}\right)}}}{4}$$

因此，

$$\int{\frac{\sqrt{57} int_{0}^{4} x^{\frac{7}{2}}}{4} d x} = \frac{\sqrt{57} int_{0}^{4} x^{\frac{9}{2}}}{18}$$

加上积分常数：

$$\int{\frac{\sqrt{57} int_{0}^{4} x^{\frac{7}{2}}}{4} d x} = \frac{\sqrt{57} int_{0}^{4} x^{\frac{9}{2}}}{18}+C$$

$$$\int \frac{\sqrt{57} int_{0}^{4} x^{2} \sqrt{x^{3}}}{4}\, dx = \frac{\sqrt{57} int_{0}^{4} x^{\frac{9}{2}}}{18} + C$$$A