Determine $\int\left[\left(\frac{1}{2}\right)^{x}+x^{1 / 2}\right] d x$
A. $\frac{1}{\ln (2) \cdot 2^{x}}+\frac{2}{3} x^{3 / 2}+C$
B. $\frac{1}{\ln (1 / 2)}\left(\frac{1}{2}\right)^{x}+\frac{2}{3} x^{3 / 2}+C$
$c-\frac{1}{\ln (x) \cdot 2^{x}}+\frac{2}{3} x^{3 / 2}+C$
D. $\frac{1}{x+1}\left(\frac{1}{2}\right)^{x+1}+\frac{2}{3} x^{3 / 2}+C$
E. $-\frac{1}{\ln (2) \cdot 2^{x}}+\frac{2}{3} x^{5 / 2}+C$

Step 1 ：The integral is a sum of two terms, so we can integrate each term separately.

Step 2 ：The first term is a simple exponential function, and the second term is a power function.

Step 3 ：The integral of an exponential function is the same function divided by the natural logarithm of its base.

Step 4 ：The integral of a power function is the function with its exponent increased by 1, divided by the new exponent.

Step 5 ：Therefore, the integral of the function is \(\frac{1}{\ln (2) \cdot 2^{x}}+\frac{2}{3} x^{3 / 2}+C\).

Step 6 ：Final Answer: \(\boxed{\frac{1}{\ln (2) \cdot 2^{x}}+\frac{2}{3} x^{3 / 2}+C}\)