[tex]I=\displaystyle \int \dfrac{\sqrt x}{\sqrt x -4} ~dx\\\\\text{Let,}\\\\~~~~u=\sqrt x-4\\\\\implies \dfrac{du}{dx} = \dfrac d{dx} \left( \sqrt x -4 \right)\\\\\implies \dfrac{du}{dx} = \dfrac 1{2 \sqrt x} \\\\\implies dx =2\sqrt x~ du\\\\\\I= \displaystyle \int \dfrac{u+4}{u+4 -4 } 2(u+4) du\\\\\\~~~=2\displaystyle \int \dfrac{(u+4)^2}{u} du\\\\\\~~~=2\displaystyle \int \dfrac{u^2+ 8u+16}{u} du\\\\\\~~~=2\displaystyle \int \left(u+8+\dfrac{16}u\right) du\\\\\\[/tex]
[tex]~~~=2\left(\displaystyle \int u ~ du + \displaystyle \int 8 ~ du + \displaystyle \int \dfrac{16} u ~ du\right) \\\\\\~~~=2\left( \dfrac {u^2}2 + 8u+16\ln |u| \right) +C\\\\\\~~~~= u^2 + 16u+ 32 \ln |u|+C\\\\\\~~~~=\left( \sqrt x -4 \right)^2 + 16\left( \sqrt x -4 \right) +32 \ln |\sqrt x -4|+C\\\\\\~~~~= x-8\sqrt x+16 +16\sqrt x -64 +32 \ln \left|\sqrt x-4 \right|+C\\\\\\\~~~~=x+8\sqrt x -48 +32\ln|\sqrt x -4| +C[/tex]
