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

Find $$$\int \frac{1}{x^{\frac{3}{2}}}\, dx$$$.

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

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

Therefore,

$$\int{\frac{1}{x^{\frac{3}{2}}} d x} = - \frac{2}{\sqrt{x}}$$

Add the constant of integration:

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

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