The total number of distinct values of x∈[0,1] for which ∫0xt21+t4dt=2x-1 is

Let fx=∫0xt21+t4dt −2x+1Clearly  f0=1,    f1=∫01t21+t4dt−1Now     t2+1t2≥2⇒t2t4+1≤12∴ f(1)<0Also   f1x=x21+x4−2=1x2+1x2−2Since  x2+1x2≥2⇒f1(x)<0∴ f is Strictly decreasing  and  f(0)>0,      f(1)<0∴ Exactly  one  root  for  f(x)  in (0,1).