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

where ln x denotes the natural logarithm of x.

DEFDBL A-Z
pi = ATN(1) * 4
h = .0000001
y0 = LOG(4 / 3)
yb = y0
CLS
   x = yb
   t = 0
   DO
     t = t + h / SQR(EXP(x) - 1)
     IF t > pi / 6 THEN EXIT DO
     tprev = t
     x = x + h
   LOOP
   PRINT x - h, tprev, tprev - pi / 6
   PRINT x, t, t - pi / 6

Finds that

    y                           integral                     diff from pi/6  
.6931469771900832           .5235987088310166          -6.676728229554297D-08
.6931470771900844           .5235988088310281           3.32327291981409D-08

so a good approximation for y would be .693147077190084, which is close enough to ln(2)~=.6931471805599453 to take the latter as probably being the answer. The discrepancy from .6931471805599453 is probably attributable to the coarseness of h in the Riemann sum.