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. $e^x \sqrt{x} + C$

Answer 1

Answer: (D)

$e^x \sqrt{x} + C$

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:

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

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:

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