Given equation is

(x+1) log (x +1) (3 + 2x – x 2 ) = (x – 3) |x| .....(i)

Clearly 3 + 2x – x 2 > 0 [log x is defined if x > 0]

⇒ x 2 – 2x – 3 < 0 ⇒ (x – 3) (x + 1) < 0

⇒ – 1 < x < 3 ...(ii)

Let x + 1 ∴ x > 0 ....(iii)

∴ (i) gives

(x+1) log x+1 (3 + 2x – x 2 ) = (x – 3)x

⇒ 3 + 2x – x 2 = x 2 – 3x

⇒ 2x 2 – 5x – 3 = 0 ⇒ x = 3, –

But from (ii), (iii) – 1 < x < 3 and x > 0

∴ x = 3, – are not possible.

Hence there is no solution

──────────────────────────────────────────────────────────────────────────────────────────