Find the following integral: $\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x$

(N/A) To evaluate the integral $\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x$,we first simplify the integrand by dividing each term in the numerator by the denominator $x^{2}$:
$\int \left( \frac{x^{3}}{x^{2}} + \frac{5x^{2}}{x^{2}} - \frac{4}{x^{2}} \right) d x$
$= \int (x + 5 - 4x^{-2}) d x$
Now,we integrate each term separately using the power rule $\int x^{n} d x = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):
$= \int x d x + \int 5 d x - \int 4x^{-2} d x$
$= \frac{x^{2}}{2} + 5x - 4 \left( \frac{x^{-2+1}}{-2+1} \right) + C$
$= \frac{x^{2}}{2} + 5x - 4 \left( \frac{x^{-1}}{-1} \right) + C$
$= \frac{x^{2}}{2} + 5x + \frac{4}{x} + C$
where $C$ is an arbitrary constant.