y
     ∫(ex - 1)-0.5 dx = pi/6 
      ln(4/3)

where ln x denotes the natural logarithm of x.

let f(x)=Sqrt[e^x - 1]

let us look at the function g(x)=ArcTan[f(x)]

taking the derivative of g we get

g'(x)=f'(x)/(1+f(x)^2)=e^x/2(e^x * f(x) )=1/2f(x)

thus we have the antiderivative of 1/f(x) is  2*ArcTan[f(x)] thus the equation becomes

2*Arctan[f(y)]-2*Arctan[f(ln(4/3))]=pi/6

2*Arctan[f(y)]-2*Arctan[sqrt(1/3)]=pi/6

2*Arctan[f(y)]-2*pi/6=pi/6

Arctan[f(y)]=pi/4

f(y)=1

e^y - 1=1

e^y=2

y=ln(2)

QED