Find the following integral: $\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$

Given the integral: $\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$
We can rewrite the integrand by dividing each term in the numerator by $\sqrt{x} = x^{\frac{1}{2}}$:
$= \int \left( \frac{x^{3}}{x^{\frac{1}{2}}} + \frac{3x}{x^{\frac{1}{2}}} + \frac{4}{x^{\frac{1}{2}}} \right) d x$
$= \int \left( x^{3-\frac{1}{2}} + 3x^{1-\frac{1}{2}} + 4x^{-\frac{1}{2}} \right) d x$
$= \int \left( x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}} \right) d x$
Using the power rule $\int x^{n} d x = \frac{x^{n+1}}{n+1} + C$:
$= \frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1} + 3 \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} + 4 \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C$
$= \frac{x^{\frac{7}{2}}}{\frac{7}{2}} + 3 \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 4 \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C$
$= \frac{2}{7} x^{\frac{7}{2}} + 2 x^{\frac{3}{2}} + 8 x^{\frac{1}{2}} + C$
$= \frac{2}{7} x^{\frac{7}{2}} + 2 x^{\frac{3}{2}} + 8 \sqrt{x} + C$,where $C$ is an arbitrary constant.