Solve the following :-

l x+1/ x l + l x + 1l = (x+1)²/ l x l

we have,

But x = – 1 is not in the interval which we are considering in this case.

So ew do not get any solution for x< -1.

When –1 ≤ x < 0, 

(x + 1)² – (x + 1) + x (x + 1) = 0

⇒(x + 1) {(x + 1) –1 + x} = 0

⇒(x + 1) × (2x) = 0

⇒x = 0 or – 1

Since x ≠ 0 ⇒ x = – 1 is a solution when – 1 ≤ x < 0.

When x ≥ 0, 

(x + 1)² – x (x + 1) – (x + 1) = 0

⇒ (x + 1) {(x + 1) – x – 1} = 0

⇒   0 = 0, which is true.

Thus for x ≥ 0 , every value of x will be the solution of given equation, provided that x ≠ 0.

Hence, the solution set of given equation is 

S = {– 1} ∪ (0, ∞)