Integral of $$$\frac{3 \sqrt{t}}{2}$$$

Find $$$\int \frac{3 \sqrt{t}}{2}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{3}{2}$$$ and $$$f{\left(t \right)} = \sqrt{t}$$$:

$${\color{red}{\int{\frac{3 \sqrt{t}}{2} d t}}} = {\color{red}{\left(\frac{3 \int{\sqrt{t} d t}}{2}\right)}}$$

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

$$\frac{3 {\color{red}{\int{\sqrt{t} d t}}}}{2}=\frac{3 {\color{red}{\int{t^{\frac{1}{2}} d t}}}}{2}=\frac{3 {\color{red}{\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{2}=\frac{3 {\color{red}{\left(\frac{2 t^{\frac{3}{2}}}{3}\right)}}}{2}$$

Therefore,

$$\int{\frac{3 \sqrt{t}}{2} d t} = t^{\frac{3}{2}}$$

Add the constant of integration:

$$\int{\frac{3 \sqrt{t}}{2} d t} = t^{\frac{3}{2}}+C$$

$$$\int \frac{3 \sqrt{t}}{2}\, dt = t^{\frac{3}{2}} + C$$$A