Answer
$y=\dfrac{x^3}{3}+\dfrac{2x^{3/2}}{3}+C$
Work Step by Step
We are given that $\dfrac{dy}{dx}=x^2+\sqrt x$
We will separate the variables to obtain:
$dy=(x^2+\sqrt x) dx$
Integrate to obtain:
$\int dy=\int(x^2+\sqrt x) dx$
This implies that $y=\int (x^2+x^{1/2}) dx=\dfrac{x^3}{3}+\dfrac{x^{3/2}}{3/2}+C$
or, $y=\dfrac{x^3}{3}+\dfrac{2x^{3/2}}{3}+C$
