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. \[ \int\left(? y^{\prime}+\frac{1}{2 \sqrt{x}} y\right) d x \] A. $\int\left(\sqrt{x} y^{\prime}+\frac{1}{2 \sqrt{x}} y\right) d x=\frac{1}{2 \sqrt{x}} y$ B. $\int\left(\sqrt{x} y^{\prime}+\frac{1}{2 \sqrt{x}} y\right) d x=y \sqrt{x}$ c. $\int\left(-\frac{1}{\frac{3}{2}} y^{\prime}+\frac{1}{2 \sqrt{x}} y\right) d x=y \sqrt{x}$ D. $\int\left(-\frac{1}{4 x^{\frac{3}{2}}} y^{\prime}+\frac{1}{2 \sqrt{x}} y\right) d x=\frac{1}{2 \sqrt{x}} y$

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 $(f(x)y(x))'$.

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

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

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

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

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

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. $\int\left(\sqrt{x} y^{\prime}+\frac{1}{2 \sqrt{x}} y\right) d x=y \sqrt{x}$.