Answer
$$
\int \frac{2^{\sqrt{x}}}{\sqrt{x}} d x= \frac{2^{\sqrt{x}+1}}{\ln 2}+C
$$
where $C$ is constant.
Work Step by Step
$$
\begin{aligned} \int \frac{2^{\sqrt{x}}}{\sqrt{x}} d x &=\int 2^{u}(2 d u) ,\quad \quad\quad \left[\begin{array}{c}{ \text {Let } \quad u=\sqrt{x}} \\ { \text {Then } \quad d u=1 /(2 \sqrt{x}) d x}\end{array}\right] \\ &=2 \cdot \frac{2^{u}}{\ln 2}+C=\frac{2^{\sqrt{x}+1}}{\ln 2}+C \end{aligned}
$$
where $C$ is constant.
