The solution set of the equation $|\frac{x+1}{x}|+|x+1|=\frac{(x+1)^2}{|x|}$, is

$\{x:x>0\}∪\{-1\}$

We have,

$|\frac{x+1}{x}|+|x+1|=\frac{(x+1)^2}{|x|}$

$⇒|\frac{x+1}{x}|+|x+1|=\frac{|x+1|^2}{|x|}$

$⇒|x+1|\left\{\begin{matrix}\frac{1}{|x|}+1-\frac{|x+1|}{|x|}\end{matrix}\right\}=0$

$⇒|x+1|=0$ or, $\frac{1}{|x|}+1=\frac{|x+1|}{|x|}$

$⇒|x+1|=0$ or, $1+|x|=|x+1|$

$⇒x=-1$ or, $1+|x|=|x+1|$

In order to solve the equation $1 +|x|=|x+1|$, we consider the following cases:

CASE I When $x <-1$

In this case, we have

$|x|=-x$ and $|x+1|=-(x+1)$

$∴1+|x|=|x+1|$

$⇒1-x=-(x+1)$, which is absurd.

CASE II When $-1≤x<0$

In this case, we have

$|x|=-x$ and $|x+1|=x+1$

$∴1+|x|=|x+1|⇒1-x=x+1⇒x=0$

But, $-1≤x<0$

So, there is no solution in this case.

CASE III When x ≥ 0.

In this case, we have

$|x|=x$ and $|x+1|=x+1$

$∴1+|x|=|x+1|$

$⇒1+x=x+1$, which is true for all x.

Clearly, the given equation is meaningful for x ≠ 0.

Hence, the solution set of the given equation is $\{x:x>0\}∪\{-1\}$.