$\int \log (2+x)^{2+x} \mathrm{~d} x=$

The integral is simplified using logarithmic properties:

$I = \int (2+x) \log (2+x) \mathrm{d} x$

Substituting $u = 2+x$ with $\mathrm{d}u = \mathrm{d}x$ yields:

$I = \int u \log u \mathrm{d} u$

Integration by parts is applied with $P = \log u$, $\mathrm{d}Q = u\mathrm{d}u$, giving $\mathrm{d}P = \frac{1}{u}\mathrm{d}u$ and $Q = \frac{u^2}{2}$:

$I = \frac{u^2}{2} \log u - \int \frac{u}{2} \mathrm{d} u$

The remaining integral evaluates to $\frac{u^2}{4}$, yielding:

$I = \frac{u^2}{2} \log u - \frac{u^2}{4} + c$

Factoring $\frac{u^2}{2}$ gives:

$I = \frac{u^2}{2} \left(\log u - \frac{1}{2}\right) + c$

Expressing $\frac{1}{2}$ as $\log \sqrt{e}$ and applying logarithmic subtraction:

$I = \frac{u^2}{2} \log \left(\frac{u}{\sqrt{e}}\right) + c$

Substituting back $u = 2+x$ produces the final result:

$I = \frac{(2+x)^2}{2} \log \left(\frac{2+x}{\sqrt{e}}\right) + c$

This matches option A.