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

$2 \sqrt{1+x}+\log \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+C$

Let

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

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

$\Rightarrow I =2 \int \frac{t^2-1+1}{t^2-1} d t=2 \int\left(1+\frac{1}{t^2-1}\right) d t$

$\Rightarrow I=2 t+\log \left|\frac{t-1}{t+1}\right|+C=2 \sqrt{1+x}+\log \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+C$