$$$\frac{4 t_{1}}{x_{1}^{2}}$$$ 关于$$$t_{1}$$$的积分
您的输入
求$$$\int \frac{4 t_{1}}{x_{1}^{2}}\, dt_{1}$$$。
解答
对 $$$c=\frac{4}{x_{1}^{2}}$$$ 和 $$$f{\left(t_{1} \right)} = t_{1}$$$ 应用常数倍法则 $$$\int c f{\left(t_{1} \right)}\, dt_{1} = c \int f{\left(t_{1} \right)}\, dt_{1}$$$：
$${\color{red}{\int{\frac{4 t_{1}}{x_{1}^{2}} d t_{1}}}} = {\color{red}{\left(\frac{4 \int{t_{1} d t_{1}}}{x_{1}^{2}}\right)}}$$
应用幂法则 $$$\int t_{1}^{n}\, dt_{1} = \frac{t_{1}^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：
$$\frac{4 {\color{red}{\int{t_{1} d t_{1}}}}}{x_{1}^{2}}=\frac{4 {\color{red}{\frac{t_{1}^{1 + 1}}{1 + 1}}}}{x_{1}^{2}}=\frac{4 {\color{red}{\left(\frac{t_{1}^{2}}{2}\right)}}}{x_{1}^{2}}$$
因此，
$$\int{\frac{4 t_{1}}{x_{1}^{2}} d t_{1}} = \frac{2 t_{1}^{2}}{x_{1}^{2}}$$
加上积分常数：
$$\int{\frac{4 t_{1}}{x_{1}^{2}} d t_{1}} = \frac{2 t_{1}^{2}}{x_{1}^{2}}+C$$
答案
$$$\int \frac{4 t_{1}}{x_{1}^{2}}\, dt_{1} = \frac{2 t_{1}^{2}}{x_{1}^{2}} + C$$$A