Area bounded by  $y=\ln (x+1),y=\ln x+1$ and their common tangent is  $\ln (e-R)-Ksq.units.$ ($R\&K$ are real numbers) the value of $R+K$ is equal to _____

$$\begin{gathered} \begin{array}{l}\begin{array}{r}y=\ln (x+1)y=\ln x+1\\ y=\ln (x+1)=\ln x+1\\ \Rightarrow \ln (x+1)-\ln x=1\end{array}\\ \Rightarrow \ln \left(\frac{x+1}{x}\right)=1\\ \Rightarrow \frac{x+1}{x}=e\end{array} \\ \Rightarrow x=\frac{1}{e-1} \end{gathered}$$

Required area 
$$\begin{gathered} =\underset{0}{\overset{\frac{1}{e-1}}{\int }}[x-\ln (x+1)dx]+\underset{\frac{1}{e-1}}{\overset{1}{\int }}[x-(\ln x+1)dx] \\ =\ln (e-1)-\frac{1}{2} \end{gathered}$$ 

R=1,k=1/2