thats okay... evaluating the integral is easy by u substitution
cant we use properties of definitive integral n substitute x+log2-x ie log2 in place of x ?
thats a good idea, lets see...
actually that wont work because the "x" in the bounds is a constant, which is "entirely different" from the variable of integration.
ohhhhh i got it now awesome thanks

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2015-04-18T14:50:41+05:30

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Substitute u=\sqrt{e^x-1}\implies du~=~\frac{e^x}{2\sqrt{e^x-1}}dx

(Notice that u=\sqrt{e^x-1}\implies u^2=e^x-1\implies 1+u^2=e^x)

therefore the earlier differential becomes du~=~\frac{1+u^2}{2\sqrt{e^x-1}}dx\implies \frac{2du}{1+u^2}=\frac{1}{\sqrt{e^x-1}}dx and the integral becomes

=\int\limits_{1}^{\sqrt{e^x-1}}\frac{2du}{1+u^2}

=2\tan^{-1}(u)\Bigg|_{1}^{\sqrt{e^x-1}}

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

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

set that equal to \frac{\pi}{6} and solve x :
2\tan^{-1}(\sqrt{e^x-1})-\frac{\pi}{2}~=~\frac{\pi}{6}
\tan^{-1}(\sqrt{e^x-1}~=~\frac{\pi}{3}
\sqrt{e^x-1}~=~\sqrt{3}
e^x-1~=~3
x~=~\log(4)
2 5 2