∫ dx/(√(5-x)) = (where C is arbitrary constant.)

$I=\int \frac{1}{\sqrt{5-x}} d x$       .....(1)

let $y=5-x$

So $d y=-d x \Rightarrow d x=-d y$

substituting y in eq. 1

$I =\int \frac{-d y}{\sqrt{y}}$

$=-\int y^{-1 / 2} d y$

$=\frac{-y^{-1 / 2+1}}{-1 / 2+1}+C=\frac{-y^{1 / 2}}{1 / 2}+C$

$=-2 \sqrt{y}+C$

substituting y in terms of x

C → arbitrary constant

$-2 \sqrt{5-x}+c$