<s>[INST] Where did I go wrong in this integration $\int\frac{\ln(1-e^x)}{e^{2x}}\,dx$ Here is the closest I've come to the answer
Link to Wolfram equality not giving true as output
NB! I marked where I'm unsure in $\color{red}{red}$ color. And please don't get startled because of the wall of equation, it's simply a more tedious integration in my opinion.
$$\int\frac{\ln(1-e^x)}{e^{2x}}\,dx$$
let: $e^x= u \iff dx=\frac{1}{u}du$
$$\int\frac{\ln(1-u)}{u^3}\,du = -\frac{\ln(1-u)}{2u^2} - \frac{1}{2}\int\frac{1}{u^2(1-u)}\,du$$
This can be written as
$$-\frac{\ln(1-u)}{2u^2}-\frac{1}{2}\int\left(\color{red}{\frac{A}{u}+\frac{B}{u^2}-\frac{C}{1-u}}\right)\,du\qquad\qquad(1)$$
where $A,B,C$ is determined through setting
$$\frac{1}{u^2(1-u)} = \color{red}{\frac{A}{u}+\frac{B}{u^2}-\frac{C}{1-u}}$$
By multiplying with LHS denominator and factoring RHS I get
$$1 = (-A-C)t^2+(A-B)t+B \implies B = A = 1 \land C = -1$$
This put into $(1)$ gives us
$$-\frac{\ln(1-u)}{2u^2}-\frac{1}{2}\int\left(\frac{1}{u}+\frac{1}{u^2}-\frac{-1}{1-u}\right)\,du$$
and
$$-\frac{1}{2}\left(\frac{\ln(1-u)}{u^2} + \ln(u) - \frac{1}{u}+\int\frac{1}{1-u}\,du\right)$$
another substitution $w = 1-u \iff du = -dw$ gives us
$$-\frac{1}{2}\left(\frac{\ln(1-u)}{u^2} + \ln(u) - \frac{1}{u}-\int\frac{1}{w}\,dw\right)$$
Finally this is equivalent to
$$-\frac{1}{2}\left(\frac{\ln(1-u)}{u^2} + \ln(u) - \frac{1}{u}-\ln(w)\right) + C$$
Substitution back to x and factoring gives us
$$-\frac{1}{2e^{2x}}\left(\ln(1-e^x) + \color{red}{e^{2x}\cdot x} - e^x - e^{2x}\ln(1-e^x)\right) + C$$
 [/INST] $$I=\int\frac{\ln(1-u)}{u^3}\,du$$
$$I=\frac{\ln(1-u)}{-2u^2}-\int\frac{1}{2u^2(1-u)}$$
Now, $$\frac{1}{u^2(1-u)}=\frac{1}{u^2}+\frac1{u}+\frac1{(1-u)}$$
So:
$$I=\frac{1}{2}\left(-\frac{\ln(1-u)}{u^2}-\left(-\frac1u+\ln u-\ln|1-u|\right)\right)+c\\
I=\frac{1}{2e^{2x}}\left(-\ln(1-e^x)+e^{x}-e^{2x}x+e^{2x}\ln|1-e^x|\right) + c$$
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