Given the integral:
\[
\int \log((2 + x)^{2 + x}) \, dx
\]
Applying the logarithmic identity $\log(a^b) = b \log a$:
\[
\log((2 + x)^{2 + x}) = (2 + x) \log(2 + x)
\]
The integral becomes:
\[
\int (2 + x) \log(2 + x) \, dx
\]
Let $t = 2 + x$, so $dt = dx$. The integral transforms to:
\[
\int t \log t \, dt
\]
Using integration by parts with $u = \log t$ and $dv = t \, dt$, we get $du = \frac{1}{t} \, dt$ and $v = \frac{t^2}{2}$.
The integration yields:
\[
\int t \log t \, dt = \frac{t^2}{2} \log t - \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt
= \frac{t^2}{2} \log t - \int \frac{t}{2} \, dt
= \frac{t^2}{2} \log t - \frac{t^2}{4} + C
\]
Substituting $t = 2 + x$ back into the result:
\[
\int \log((2 + x)^{2 + x}) \, dx = \frac{(2 + x)^2}{2} \log(2 + x) - \frac{(2 + x)^2}{4} + C
\]
This can also be expressed as:
\[
(2 + x)(2 + x)^x (\log(2 + x) + 1) + C
\]