Integral of $$$- \frac{3}{\sqrt{y^{3}}}$$$

Find $$$\int \left(- \frac{3}{\sqrt{y^{3}}}\right)\, dy$$$.

The input is rewritten: $$$\int{\left(- \frac{3}{\sqrt{y^{3}}}\right)d y}=\int{\left(- \frac{3}{y^{\frac{3}{2}}}\right)d y}$$$.

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

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

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

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

Therefore,

$$\int{\left(- \frac{3}{y^{\frac{3}{2}}}\right)d y} = \frac{6}{\sqrt{y}}$$

Add the constant of integration:

$$\int{\left(- \frac{3}{y^{\frac{3}{2}}}\right)d y} = \frac{6}{\sqrt{y}}+C$$

$$$\int \left(- \frac{3}{\sqrt{y^{3}}}\right)\, dy = \frac{6}{\sqrt{y}} + C$$$A