Integral of $$$\frac{4}{x^{\frac{3}{4}}}$$$

Find $$$\int \frac{4}{x^{\frac{3}{4}}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \frac{1}{x^{\frac{3}{4}}}$$$:

$${\color{red}{\int{\frac{4}{x^{\frac{3}{4}}} d x}}} = {\color{red}{\left(4 \int{\frac{1}{x^{\frac{3}{4}}} d x}\right)}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{3}{4}$$$:

$$4 {\color{red}{\int{\frac{1}{x^{\frac{3}{4}}} d x}}}=4 {\color{red}{\int{x^{- \frac{3}{4}} d x}}}=4 {\color{red}{\frac{x^{- \frac{3}{4} + 1}}{- \frac{3}{4} + 1}}}=4 {\color{red}{\left(4 x^{\frac{1}{4}}\right)}}=4 {\color{red}{\left(4 \sqrt[4]{x}\right)}}$$

Therefore,

$$\int{\frac{4}{x^{\frac{3}{4}}} d x} = 16 \sqrt[4]{x}$$

Add the constant of integration:

$$\int{\frac{4}{x^{\frac{3}{4}}} d x} = 16 \sqrt[4]{x}+C$$

$$$\int \frac{4}{x^{\frac{3}{4}}}\, dx = 16 \sqrt[4]{x} + C$$$A