When solving integrals with mixed functions like exponentials and rational functions, it is helpful to split the integral into separate parts. Look for substitution opportunities (like \( u = \sqrt{x} \)) to simplify the integrand. Additionally, remember that the exponential function, \( e^x \), integrates easily, and such terms typically appear straightforwardly in the result. Always verify your substitution and simplify step-by-step.
The integral to be evaluated is:
\[ I = \int e^x \left( \frac{2x + 1}{2\sqrt{x}} \right) dx. \]
The integrand simplifies as follows:
\[ \frac{2x + 1}{2\sqrt{x}} = \frac{2x}{2\sqrt{x}} + \frac{1}{2\sqrt{x}} = \sqrt{x} + \frac{1}{2\sqrt{x}}. \]
Substituting this simplification into the integral yields:
\[ I = \int e^x \left( \sqrt{x} + \frac{1}{2\sqrt{x}} \right) dx. \]
The integral can be split into two parts:
\[ I = \int e^x \sqrt{x} \, dx + \frac{1}{2} \int e^x \frac{dx}{\sqrt{x}}. \]
Using the substitution \( u = \sqrt{x} \), which implies \( x = u^2 \) and \( dx = 2u \, du \), we can transform both terms.
For the first integral term:
\[ \int e^x \sqrt{x} \, dx = \int e^x u \cdot 2u \, du = \int e^x u^2 \, du. \]
Applying the substitution \( u = \sqrt{x} \), this term becomes \( e^x u^2 = e^x \sqrt{x}. \)
For the second integral term:
\[ \frac{1}{2} \int e^x \frac{dx}{\sqrt{x}} = \frac{1}{2} \int e^x u^{-1} \cdot 2u \, du = \int e^x du = e^x. \]
Combining the results of both terms:
\[ I = e^x \sqrt{x} + e^x + C. \]
The final simplified form of the integral is:
\[ I = e^x \sqrt{x} + C. \]