√(x−3)(2−x)<√4x2+12x+11
For the square root to exist,
(x−3)(2−x)≥0⇒(x−3)(x−2)≤0⇒x∈[2,3]
4x2+12x+11≥0⇒4x2+12x+11≥0⇒D=144−4×44<0∴4x2+12x+11>0 ∀ x∈R
So, x∈[2,3]
Now, squaring on both sides
−x2+5x−6<4x2+12x+11⇒5x2+7x+17>0D=49−4×5×17<0∴5x2+7x+17>0 ∀ x∈R
Therefore, x∈[2,3]
There are 2 integral values in the range of x.