Question

\(\text{Evaluate } \int e^x \left( \frac{2x + 1}{2 \sqrt{x}} \right) dx:\)

• A. \( \frac{1}{2\sqrt{x}} e^x + C \)
• B. \( -e^x \sqrt{x} + C \)
• C. \( -\frac{1}{2\sqrt{x}} e^x + C \)
• D. \( e^x \sqrt{x} + C \)

Hint:

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.

Answer 1

The Correct Option is D

The given integral is:

\[ I = \int e^x \left( \frac{2x + 1}{2\sqrt{x}} \right) dx. \]

Simplify the integrand:

\[ \frac{2x + 1}{2\sqrt{x}} = \frac{2x}{2\sqrt{x}} + \frac{1}{2\sqrt{x}} = \sqrt{x} + \frac{1}{2\sqrt{x}}. \]

Substitute this into the integral:

\[ I = \int e^x \left( \sqrt{x} + \frac{1}{2\sqrt{x}} \right) dx. \]

Split the integral:

\[ I = \int e^x \sqrt{x} \, dx + \frac{1}{2} \int e^x \frac{dx}{\sqrt{x}}. \]

Let \( u = \sqrt{x} \), so \( x = u^2 \) and \( dx = 2u \, du \). Substitute into both terms.

For the first term:

\[ \int e^x \sqrt{x} \, dx = \int e^x u \cdot 2u \, du = \int e^x u^2 \, du = e^x u^2 = e^x \sqrt{x}. \]

For the second term:

\[ \frac{1}{2} \int e^x \frac{dx}{\sqrt{x}} = \frac{1}{2} \int e^x u^{-1} \cdot 2u \, du = \int e^x du = e^x. \]

Combine the results:

\[ I = e^x \sqrt{x} + e^x + C. \]

Thus:

\[ I = e^x \sqrt{x} + C. \]

Answer 2

The given integral is: 

\[ I = \int e^x \left( \frac{2x + 1}{2\sqrt{x}} \right) dx. \]

Simplify the integrand:

\[ \frac{2x + 1}{2\sqrt{x}} = \frac{2x}{2\sqrt{x}} + \frac{1}{2\sqrt{x}} = \sqrt{x} + \frac{1}{2\sqrt{x}}. \]

Substitute this into the integral:

\[ I = \int e^x \left( \sqrt{x} + \frac{1}{2\sqrt{x}} \right) dx. \]

Split the integral:

\[ I = \int e^x \sqrt{x} \, dx + \frac{1}{2} \int e^x \frac{dx}{\sqrt{x}}. \]

Substitute \( u = \sqrt{x} \), so \( x = u^2 \) and \( dx = 2u \, du \):

For the first term:

\[ \int e^x \sqrt{x} \, dx = \int e^x u \cdot 2u \, du = \int e^x u^2 \, du = e^x u^2 = e^x \sqrt{x}. \]

For the second term:

\[ \frac{1}{2} \int e^x \frac{dx}{\sqrt{x}} = \frac{1}{2} \int e^x u^{-1} \cdot 2u \, du = \int e^x du = e^x. \]

Combine the results:

\[ I = e^x \sqrt{x} + e^x + C. \]

Thus:

\[ I = e^x \sqrt{x} + C. \]