Integral of $$$- \frac{1}{2 x^{\frac{3}{2}}}$$$

Find $$$\int \left(- \frac{1}{2 x^{\frac{3}{2}}}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{1}{2}$$$ and $$$f{\left(x \right)} = \frac{1}{x^{\frac{3}{2}}}$$$:

$${\color{red}{\int{\left(- \frac{1}{2 x^{\frac{3}{2}}}\right)d x}}} = {\color{red}{\left(- \frac{\int{\frac{1}{x^{\frac{3}{2}}} d x}}{2}\right)}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{3}{2}$$$:

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

Therefore,

$$\int{\left(- \frac{1}{2 x^{\frac{3}{2}}}\right)d x} = \frac{1}{\sqrt{x}}$$

Add the constant of integration:

$$\int{\left(- \frac{1}{2 x^{\frac{3}{2}}}\right)d x} = \frac{1}{\sqrt{x}}+C$$

$$$\int \left(- \frac{1}{2 x^{\frac{3}{2}}}\right)\, dx = \frac{1}{\sqrt{x}} + C$$$A