Solve the equation $$(x+4)3^{1-|x-1|}-x=(x+1)|3^x-1|+3^{x+1}+1$$
As I don't see anything else, my idea was to see where the roots of each of the absolute values are, so: $$x-1=0\Rightarrow 1\\3^x-1=0\iff 3^x=1\iff x=0$$ Then we can look at the intervals $R_1=(-\infty;0),R_2=[0;1)$ and $R_3=[1;+\infty)$ in each of which we can get rid of the absolute values.
In $R_1=(-\infty;0)\Rightarrow$ $$3^x(x+4)-x=(x+1)(1-3^x)+3^{x+1}+1\\4.3^x+3^x-3^{x+1}=x-2x3^x+2\\3^x(4+1-3)=x-2x3^x+2$$ I am not sure how to proceed as we have the unknown outside of the exponents.
As commented by Sil, from $$3^x(4+1-3)=\color{red}2x-2x3^x+2$$ one can get $$2\cdot 3^x-2x+2x3^x-2=0$$ $$3^x(2+2x)-(2+2x)=0$$ $$(2+2x)(3^x-1)=0$$ from which $x=-1$ follows.
If $0\leqslant x\lt 1$, one can see that the equation always holds.
If $1\leqslant x$, one has $$(x+4)3^{2-x}-x=(x+1)(3^x-1)+3^{x+1}+1$$ $$(x+1)3^x+3^{x+1}-(x+4)3^{2-x}=0$$ $$(x+4)(3^x-3^{2-x})=0$$ from which $x=1$ follows.
Another way :
The equation can be written as $$3^x(x+4)(\underbrace{3^{-x+1-|x-1|}-1}_{A})+(x+1)(\underbrace{3^x-1-|3^x-1|}_{B})=0$$ from which one can get the answer using that $A=0\iff x\leqslant 1$, and that $B=0\iff x\geqslant 0$.