If $\int\frac{1}{x}\log_{ex}e.\log_{e^2x}e\,dx=\log(\frac{1+f(x)}{2+f(x)})+C$ then $\underset{x→0}{\lim}\frac{f(1+x)}{x}$ equals:

1

$\int\frac{1}{x}\log_{ex}e.\log_{e^2x}e\,dx=\int\frac{1}{x}(\frac{\log_e}{\log_e+\log_x})dx=\int\frac{1}{x}(\frac{\log_e}{2\log_e+\log_x})dx=\int\frac{1}{x}\frac{1}{(1+\log x)}\frac{1}{(2+\log x)}dx$

Let log x = k $⇒\int\frac{1}{e^k}\frac{1}{(1+k)}\frac{1}{(2+k)}e^k\,dk=\int(\frac{1}{1+k}\frac{1}{2+k})dk=\log(\frac{1+k}{2+k})$

$⇒f(x)=k=\log x⇒\underset{x→0}{\lim}\frac{\log(x+1)}{x}(\frac{0}{0}form)$

⇒ from L.H. rule $\underset{x→0}{\lim}\frac{1}{x+1}=1$