Integral of $$$\frac{23}{50 t^{2}}$$$

Find $$$\int \frac{23}{50 t^{2}}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{23}{50}$$$ and $$$f{\left(t \right)} = \frac{1}{t^{2}}$$$:

$${\color{red}{\int{\frac{23}{50 t^{2}} d t}}} = {\color{red}{\left(\frac{23 \int{\frac{1}{t^{2}} d t}}{50}\right)}}$$

Apply the power rule $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

$$\frac{23 {\color{red}{\int{\frac{1}{t^{2}} d t}}}}{50}=\frac{23 {\color{red}{\int{t^{-2} d t}}}}{50}=\frac{23 {\color{red}{\frac{t^{-2 + 1}}{-2 + 1}}}}{50}=\frac{23 {\color{red}{\left(- t^{-1}\right)}}}{50}=\frac{23 {\color{red}{\left(- \frac{1}{t}\right)}}}{50}$$

Therefore,

$$\int{\frac{23}{50 t^{2}} d t} = - \frac{23}{50 t}$$

Add the constant of integration:

$$\int{\frac{23}{50 t^{2}} d t} = - \frac{23}{50 t}+C$$

$$$\int \frac{23}{50 t^{2}}\, dt = - \frac{23}{50 t} + C$$$A