The value of $f\int \frac{\sqrt{1+x}}{x}dx$, is

Solution:

Let 

$I=\int \frac{\sqrt{1+x}}{x}dx$

$\Rightarrow I=\int \frac{\sqrt{t^2}}{t^2-1}-2tdt,$ where $1+x=t^2$

$\begin{array}{l}\Rightarrow t=2\int \frac{t^2-1+1}{t^2-1}dt=2\int \left(1+\frac{1}{t^2-1}\right)dt\\ \Rightarrow I=2t+\log \left|\frac{t-1}{t+1}\right|+C=2\sqrt{1+x}+\log \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+C\end{array}$