From Schwinger's paper, chapter 6, this expression for the operator $U(s)=e^{-i(H_0+H_1)s}$ can be easily obtained (the commutator $[H_0, H_1]$ is unuseful, just know it's not 0). So
$U(s)=U_0(s)-is\int_{0}^{1}du_1U_0(s(1-u_1))H_1U_0(su_1)+\\ +(-is)^2\int_{0}^{1}du_1U_0(s(1-u_1))H_1U_0(u_1s)\int_{0}^{1}u_1du_2U_0(su_1(1-u_2))H_1U_0(su_1u_2)+\\ +\ldots\\ +(-is)^n\int_{0}^{1}u_1^{n-1}du_1\int_{0}^{1}u_2^{n-2}du_2\ldots\int_{0}^{1}du_n \times\Bigg [ U_0(s(1-u_1))H_1U_0(su_1)\ldots U_0(su_1\ldots u_{n-1}(1-u_n))H_1U_0(su_1\ldots u_n)\Bigg ]+\ldots$
where $U_0(s)=e^{-iH_0s}$. From here, he writes the integral (which is only valid under the trace Tr)
$\text{Tr}\;U(s)=\text{Tr}\;U_0(s)-is\int_{0}^{1}d\lambda\;\text{Tr}\;H_1e^{-i(H_0+\lambda H_1)s}$
(I can understand its derivation, but cannot prove it step by step)
and says: "we insert the first expression ($U(s)$) for $e^{-i(H_0+\lambda H_1)s}$" to obtain
$\text{Tr}\;U(s)=\text{Tr}\;U_0(s)-is\text{Tr}\;H_1U_0(s)]+\\+\frac{1}{2}(-is)^2\int_0^1du_1\;\text{Tr}\;[H_1U_0((1-u_1)s)H_1U_0(u_1s)]+\ldots+\\ +\frac{(-is)^{n+1}}{n+1}\int_0^1u_1^{n-1}du_1\ldots\int_0^1du_n\;\text{Tr}\;[H_1U_0((1-u_1)s)H_1\ldots H_1(u_1\ldots u_ns)]+\ldots$
Now, I have tried using a simple expansion for the exponential, a "reverse" Baker Campbell Hausdorff formula (just swap $B$ for $B$ minus the commutators), the cyclic property of the trace, and nothing seems to really work. The closest thing is plugging in $e^{A+B}=e^Ae^{B-\frac{1}{2}[A,B]+..}$ with $A=-iH_0s$ and $B=-i\lambda H_1s$ to obtain
$\text{Tr}\;U(s)=\text{Tr}\;U_0(s)-is\int_{0}^{1}d\lambda\;\text{Tr}\;H_1e^{-i(H_0s)}\exp{\left((-is)\lambda H_1-(-is)^2\frac{\lambda^2}{2}[H_0,H_1]+\ldots\right)}$
from which nothing useful comes out. The main issue seems to be the fact that there needs to be a substitution in the first expression to conserve the $du_i$ integrals, but from which integration in $d\lambda$ gives the factors in the denominators ($n+1$ etc). A suggestion I got from a professor is to try and use the cyclic property of the trace to make the commutators vanish, but I am not sure how. A simple yet very informal way would be to just integrate the first expression in $d\lambda$ with $\lambda H_1$ in place of $H_1$.
Any ideas?