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

where ln x denotes the natural logarithm of x.

Hi!

I found the same result y=ln(2) by switching to other integral like this:

x=ln(t^2+1)

So dx=2*t/(t^2+1)dt

and x1=ln(4/3) => t1=1/sqrt(3)

      x2=y => t2=sqrt(e^y-1)

The new integral is 2arctan(t) from t1 to t2.

After some calculus i receive y=ln(2)