Concept:         

Integration formulas:

• \(\int_a^b {f(x)dx = F(b) - F(a)} \)
• \(\int_a^b {\frac{1}{{1 + {x^2}}}dx = } {\tan ^{ - 1}}x\)

Calculation:      

Let us suppose, e^x - 1 = t²

Now by differentiating the above equation w.r.t x we get

⇒ e^x dx = 2t dt

\(\Rightarrow dx = \frac{{2t\ dt}}{{{t^2} + 1}}\)

Put the value of e^x-1 and dx in given integration

\(\int {\frac{1}{{\sqrt {{t^2}} }}} \times \frac{{2tdt}}{{{t^2} + 1}}\)

\(\int 2 \frac{{dt}}{{{t^2} + 1}} ⇒ 2{\tan ^{ - 1}}t\)

Now put the value of t and limit in the above integrand

\([2{\tan ^{ - 1}}\sqrt {{e^x} - 1} ]_{\log 2}^x = \frac{\pi }{6}\)

\(2{\tan ^{ - 1}}\sqrt {{e^x} - 1} - 2{\tan ^{ - 1}}\sqrt {{e^{\log 2}} - 1} = \frac{\pi }{6}\)

\(2{\tan ^{ - 1}}\sqrt {{e^x} - 1} - 2{\tan ^{ - 1}}1 = \frac{\pi }{6}\)

\(2{\tan ^{ - 1}}\sqrt {{e^x} - 1} - \frac{\pi }{2} = \frac{\pi }{6}\)

\(2{\tan ^{ - 1}}\sqrt {{e^x} - 1} = \frac{{2\pi }}{3}\)

\(\sqrt {{e^x} - 1} = \tan \frac{\pi }{3} = \sqrt 3 \)

\({e^x} - 1 = 3 ⇒ {e^x} = 4 ⇒ x = 2\log 2\)

Hence, option B is the correct answer.