The integral $\int \frac{dx}{|1+\sqrt{x}|-\sqrt{x^2}}$ is equal to

Put, $1+\sqrt{x}=t \Rightarrow \frac{1}{2\sqrt{x}}dx=dt$

And, $1-x=1-{\left(t-1\right)}^2=2t-t^2$

$\therefore I= \int \frac{2dt}{t\sqrt{2t-t^2}}$

Again put, $t=\frac{1}{z}\Rightarrow dt=\frac{-1}{z^2}dz$

$\Rightarrow I=2\int \frac{-\frac{1}{z^2} dz}{\frac{1}{z} \sqrt{\frac{2}{z}-\frac{1}{z^2}}}=2\int \frac{-dz}{\sqrt{2z-1}}=-2\sqrt{2z-1}+c$

Where $c$ is arbitrary constant.

$=-2\sqrt{\frac{2}{t}-1 }+c$

$=-2\sqrt{\frac{2-t}{t}}+c$

$=-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+c$