An interesting integral, I don't know if it can be calculated, but we can try using a binomial series:
$$ (1+x)^{x}=\sum_{k=0}^{\infty}\left(\begin{array}{l} x \\ k \end{array}\right) x^{k} $$
And then integrate.
Maple gave me $(n+1)^n-\frac{1}{n+1}$. But Wolfram breaks this "theory" already for $n=1$. Perhaps this is asymptotic behavior for large $n$.
What you could do is a series expansion $$(1+x)^x=1+\sum_{k=2}^\infty (-1)^k\,\frac{a_k}{k!}\, x^k$$ where the first coefficients $a_k$ are $$\{2,3,20,90,594,4200,34544,316008,3207240,35699400,432690312,\cdots\}$$ This is sequence $A066166$ in $OEIS$.
As you can see in the formula section, in year $2014$ Vaclav Kotesovec proposed, as an approximation, $$a_k \sim k! \left(1-\frac{1}{k}-\frac{\log (k)+\gamma -1}{k^2}\right)$$
So, for the integral, $$\int_0^n (1+x)^x\, dx=n+\sum_{k=2}^\infty (-1)^k\frac {a_k}{(k+1)!}\, n^{k+1}$$ whjci will grow faster than an exponential.
By L'Hôpital's rule, $$ \mathop {\lim }\limits_{n \to + \infty } \frac{{\int_0^n {(1 + x)^x dx} }}{{\frac{{(1 + n)^n }}{{\log (1 + n)}}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{(1 + n)^n }}{{(1 + n)^n \left( {1 + \frac{n}{{(1 + n)\log (1 + n)}} - \frac{1}{{(1 + n)\log ^2 (1 + n)}}} \right)}} = 1, $$ which leads to the asymptotic formula $$ \int_0^n {(1 + x)^x dx} \sim \frac{{(1 + n)^n }}{{\log (1 + n)}} $$ as $n\to +\infty$. With a more sophisticated application of L'Hôpital's rule, one can show the considerably more precise asymptotic formula $$ \int_0^n {(1 + x)^x dx}=\frac{{(1 + n)^n }}{{\log (1 + n) + 1}}\left( {1 + \frac{{1 + o(1)}}{{(1 + n)\log (1 + n)}}} \right) $$ as $n\to +\infty$.
As I have stated in a comment above, the binomial series only works when $|x|<1$. This means that for higher $n$ (nonzero in this case since $n \in \mathbb{N} $), the simplification to summation no longer works, so the integral gives an incorrect value.
Some resources on binomial series:
Using this graph with verifiable results, we can solve the following. There may be a minor typo in here, but I tried as hard as I could to be accurate. I used basic Binomial Theorem, the Maclaurin Series for exp(z), the upper and lower incomplete gamma function. I skipped a bit using the first graph link and this integration source to simplify:
$$I(n)=\int_0^n (x+1)^xdx =\int_1^{n+1}x^{x-1}dx=\quad\sum_{m=0}^\infty \sum_{k=0}^m \frac{(-1)^{m-k}}{k!(m-k)!} (-(k+1))^{-(m+1)} \big(Γ(m+1)-Γ(m+1,-(k+1)\ln(n+1))\big)=\quad \sum_{m=1}^\infty \sum_{k=1}^{m-1}\frac{(-1)^kγ(m,-k\ln(n))}{Γ(k)(m-k)!k^m}$$
This may be able to be simplified like here, but the alternating part is simpler than the answer’s. Please correct me and give me feedback!
