Solving $\int_0^1 \frac{dx}{e^x-e^{-2x}+2}$ with substitution $$\int_0^1 \frac{dx}{e^x-e^{-2x}+2}$$
I've tried solve this with substitution, first with $ u = e^x$ witch gives me $\int_0^1 \frac{1}{u\left(1 - \frac{1}{e^2}\right)+2} du$ and second with $ u = e^x-e^{-2x}$ witch gives me $\int_0^1\frac{u}{u - \frac{u}{e^2}+2-\frac{2}{e^2}}du$. None of these seems any good to continue with.
== Edit ==
Forget a +2 term in the first statement  (that's why the substitutions seemed wrong (or at least a part of why))

The OP now modified the question to take into account the remark made at the end of this answer. I leave this answer as it is.
The substitution $u=\mathrm e^x$ yields $\mathrm dx=\frac1u\mathrm du$ and $\mathrm e^{-2x}=\frac1{u^2}$ hence the integral to be computed is
$$
\int_1^\mathrm e\frac{\mathrm du}{u\cdot(u-\frac1{u^2})}=\int_1^\mathrm e\frac{u\cdot\mathrm du}{u^3-1},
$$
not what you wrote. Can you continue?
By the way, the integral diverges, that is,
$$
\int_0^1\frac{\mathrm dx}{\mathrm e^x-\mathrm e^{-2x}}=+\infty.
$$
