Question: > 
> If $\int^x\_{\log 2}\dfrac{dy}{e^y-1}=\dfrac{\pi}{6}$, then find x
> 
> 
> 

Set $e^y-1=t\implies e^y.dy=dt$
$$
\int^x\_{\log 2}\dfrac{dy}{e^y-1}=\int^x\_{\log 2}\dfrac{e^y.dy}{e^y(e^y-1)}=\int\_1^{e^x-1}\frac{dt}{t(t+1)}=\int\_1^{e^x-1}\Big[\frac{1}{t}-\frac{1}{t+1}dt\Big]\\
=\bigg[\log\dfrac{t}{t+1}\bigg]\_1^{e^x-1}=\log\frac{e^x-1}{e^x}-\log\frac{1}{2}=\frac{\pi}{6}
$$

I think something is not right, the solution given in my reference is $x=\log 4$, so what's going wrong with my attempt here ?

Answer: 
Let $q=e^{-x}$.
Then
$$\begin{align}
\ln(1-q)&=\frac\pi6-\ln2\\
1-q&=\frac12\exp\frac\pi6\\
q&=1-\frac12\exp\frac\pi6\\
\Rightarrow x&=-\ln\left(1-\frac12\exp\frac\pi6\right).
\end{align}$$
$$I=\int\_{\ln 2}^{x} \frac{dy}{e^y-1} =\ln(1-e^{-x})-\ln (1/2)=\frac{\pi}{6} \implies x=-\ln(1-\frac{e^{-\pi/6}}{2}).$$