What is $\int e^x\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)dx$ equal to?

We use the standard integral form:
$$\int e^x[f(x)+f^{\prime }(x)]dx=e^{x}f(x)+C$$
In this problem, let $f(x)=\sqrt{x}$.Then $f^{\prime }(x)=\frac{d}{dx}(x^{1/2})=\frac{1}{2}{x}^{-1/2}=\frac{1}{2\sqrt{x}}$.The integral is in the form $\int e^x[f(x)+f^{\prime }(x)]dx$.Thus, the result is $e^{x}f(x)+C=e^x\sqrt{x}+C$.