√log2x−1−12log2(x3)+2>0x>0 and log2x−1≥0⇒x>0 and x≥2
⇒√log2x−1−32log2x+2>0
⇒√log2x−1−32(log2x−1)+12>0
Let
√log2x−1=t≥0
⇒t−32t2+12>0
⇒3t2−2t−1<0⇒(3t+1)(t−1)<0
⇒−13<t<1
So,
0≤t<1
⇒0≤√log2x−1<1
⇒0≤log2x−1<1
⇒1≤log2x<2
⇒2≤x<4
Hence, the integral values are 2,3
Sum =5