Given a n by n matrix $A$ which is diagonalizable and $a_{ii} = 1$ $\forall i$, I was wondering if the following calculation is possible with trace, \begin{equation} tr \int_0^1 e^{tA} dt = \int_0^1 tr(e^{tA}) dt \end{equation}
We perform an eigenvalue decomposition and get, $tr(Qe^{t\Lambda}Q^{-1})= tr(e^{t\Lambda}) = \sum\limits_{i=0}^n e^{t\lambda_i}$ $\Rightarrow$
$$\int_0^1 \sum\limits_{i=0}^n e^{t\lambda_i} dt= \sum\limits_{i=0}^n\int_0^1 e^{t\lambda_i} dt= \sum\limits_{i=0}^n \frac{e^{\lambda_i} - 1}{\lambda_i} $$ I'm not confidant about the first equality (bringing trace inside integral)
As mentioned in the comments, the assertion "$\operatorname{tr}(e^{tA}) = e^{\operatorname{tr}(tA)}$" is simply false.
On the other hand, the integration problem is straightforward. We have $\exp(tA) = \sum_{n\geq 0} A^n \frac{t^n}{n!}$ for any finite-dimensional matrix $A$, since $\exp$ has infinite radius of convergence. By linearity of integration, and checking that certain limits are sufficiently uniform, we have:
$$ \operatorname{tr} \int_0^1 e^{tA}\,\mathrm{d}t = \operatorname{tr} \int_0^1 \sum_{n\geq 0} A^n\, \frac{t^n}{n!}\,\mathrm{d}t = \operatorname{tr}\sum_{n\geq 0} A^n \int_0^1 \frac{t^n}{n!} \,\mathrm d t = \operatorname{tr}\sum_{n\geq 0} \frac{A^n}{(n+1)!} = \operatorname{tr}\left(\frac{e^A-1}{A}\right).$$