$\int \mathrm{e}^{\mathrm{x}}\left(\frac{2 \mathrm{x}+1}{2 \sqrt{\mathrm{x}}}\right) \mathrm{dx}$

First, let's simplify the expression:

$\int e^x \left(\dfrac{2x+1}{2\sqrt{x}}\right) dx = \int e^x \cdot \dfrac{2x+1}{2} \cdot x^{-1/2} dx$

$= \int e^x \cdot \left(\dfrac{2x}{2} \cdot x^{-1/2} + \dfrac{1}{2} \cdot x^{-1/2}\right) dx$

$= \int e^x \cdot \left(x^{1/2} + \dfrac{1}{2} \cdot x^{-1/2}\right) dx$

Splitting the integral:

$\int e^x \cdot \left(x^{1/2} + \dfrac{1}{2} \cdot x^{-1/2}\right) dx = \int e^x \cdot x^{1/2} dx + \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx$

Using integration by parts for the first part with:

$u = x^{1/2}$ so $du = \dfrac{1}{2} \cdot x^{-1/2} dx$

$dv = e^x dx$ so $v = e^x$

$\int e^x \cdot x^{1/2} dx = x^{1/2} \cdot e^x - \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx$

Notice that the second term is the negative of our second integral:

$\int e^x \cdot x^{1/2} dx + \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx = x^{1/2} \cdot e^x - \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx + \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx$

The last two integrals cancel out, giving us:

$\int e^x \left(\dfrac{2x+1}{2\sqrt{x}}\right) dx = \sqrt{x} \cdot e^x + C$