$$$\frac{4 t_{1}}{x_{1}^{2}}$$$ 對 $$$t_{1}$$$ 的積分

求$$$\int \frac{4 t_{1}}{x_{1}^{2}}\, dt_{1}$$$。

套用常數倍法則 $$$\int c f{\left(t_{1} \right)}\, dt_{1} = c \int f{\left(t_{1} \right)}\, dt_{1}$$$，使用 $$$c=\frac{4}{x_{1}^{2}}$$$ 與 $$$f{\left(t_{1} \right)} = t_{1}$$$：

$${\color{red}{\int{\frac{4 t_{1}}{x_{1}^{2}} d t_{1}}}} = {\color{red}{\left(\frac{4 \int{t_{1} d t_{1}}}{x_{1}^{2}}\right)}}$$

套用冪次法則 $$$\int t_{1}^{n}\, dt_{1} = \frac{t_{1}^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：

$$\frac{4 {\color{red}{\int{t_{1} d t_{1}}}}}{x_{1}^{2}}=\frac{4 {\color{red}{\frac{t_{1}^{1 + 1}}{1 + 1}}}}{x_{1}^{2}}=\frac{4 {\color{red}{\left(\frac{t_{1}^{2}}{2}\right)}}}{x_{1}^{2}}$$

因此，

$$\int{\frac{4 t_{1}}{x_{1}^{2}} d t_{1}} = \frac{2 t_{1}^{2}}{x_{1}^{2}}$$

加上積分常數：

$$\int{\frac{4 t_{1}}{x_{1}^{2}} d t_{1}} = \frac{2 t_{1}^{2}}{x_{1}^{2}}+C$$

$$$\int \frac{4 t_{1}}{x_{1}^{2}}\, dt_{1} = \frac{2 t_{1}^{2}}{x_{1}^{2}} + C$$$A