Replace the ? with a function of $x$ that will make the integrand equal to the derivative of a product, and then find the antiderivative. Choose 0 for the constant of integration.
$$\int (?y^{\prime }+\frac{1}{2\sqrt{x}}y)dx$$
A. $\int (\sqrt{x}{y}^{\prime }+\frac{1}{2\sqrt{x}}y)dx=\frac{1}{2\sqrt{x}}y$
B. $\int (\sqrt{x}{y}^{\prime }+\frac{1}{2\sqrt{x}}y)dx=y\sqrt{x}$
c. $\int (-\frac{1}{\frac{3}{2}}{y}^{\prime }+\frac{1}{2\sqrt{x}}y)dx=y\sqrt{x}$
D. $\int (-\frac{1}{4x^{\frac{3}{2}}}{y}^{\prime }+\frac{1}{2\sqrt{x}}y)dx=\frac{1}{2\sqrt{x}}y$

Step 1 ：First, we need to find a function of $x$ that will make the integrand equal to the derivative of a product. This means we need to find a function $f(x)$ such that the integrand is equal to $(f(x)y(x){)}^{\prime }$.

Step 2 ：We know that the derivative of a product of two functions is given by the product rule: $(f(x)y(x){)}^{\prime }=f^{\prime }(x)y(x)+f(x)y^{\prime }(x)$.

Step 3 ：Comparing this with the integrand, we see that $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$ and $f(x)=?$.

Step 4 ：We can find $f(x)$ by integrating $f^{\prime }(x)$: $f(x)=\int f^{\prime }(x)dx=\int \frac{1}{2\sqrt{x}}dx=\sqrt{x}$.

Step 5 ：So, the function that replaces the question mark is $\sqrt{x}$.

Step 6 ：Now, we can write the integrand as $(\sqrt{x}y(x){)}^{\prime }$ and find the antiderivative: $\int (\sqrt{x}y(x){)}^{\prime }dx=\sqrt{x}y(x)$.

Step 7 ：Since we are asked to choose 0 for the constant of integration, the final answer is $\sqrt{x}y(x)$.

Step 8 ：Therefore, the correct answer is B. $\int (\sqrt{x}{y}^{\prime }+\frac{1}{2\sqrt{x}}y)dx=y\sqrt{x}$.