int sqrt((5-x)/(5+x)) dx Use integration tables to find the indefinite integral.

Indefinite integral follows the formula: int f(x) dx = F(x)+C
where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as constant of integration.
 The given integral problem: int sqrt ((5-x)/(5+x))dxresembles one of the formulas from the integration table. It follows the integration formula for rational function with roots as:
int sqrt(x/(a-x)) =-sqrt(x(a-x)) - a* arctan(sqrt(x(a-x))/(x-a))+C
For easier comparison, we may apply u-substitution by letting: u =5-x rearrange into x = 5-u .
The derivative of u will be du = -1 dx rearrange into -du = dx .
Plug -in the value on the integral problem, we get:
int sqrt ((5-x)/(5+x)) dx =int sqrt (u/(5+(5-u)) )* (-du)
                        =int -sqrt (u/(5+5-u)) du
                        =int -sqrt (u/(10-u)) du
Apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int -sqrt (u/(10-u)) du=(-1)int sqrt (u/(10-u)) du
By comparing "a-x " with "10-u ", we determine the corresponding value: a=10 .
Applying the aforementioned formula for rational function with roots, we get:
(-1)int sqrt (u/(10-u)) du = (-1) *[-sqrt(u(10-u)) - 10* arctan(sqrt(u(10-u))/(u-10))]+C
            =sqrt(u(10-u)) + 10* arctan(sqrt(u(10-u))/(u-10))+C
Plug-in u =5-x on sqrt(u(10-u)) + 10* arctan(sqrt(u(10-u))/(u-10))]+C , we get the indefinite integral as:
int sqrt ((5-x)/(5+x)) dx =sqrt((5-x)(10-(5-x))) + 10* arctan(sqrt((5-x)(10-(5-x)))/((5-x)-10))+C
=sqrt((5-x)(10-5+x)) + 10* arctan(sqrt((5-x)(10-5+x))/(5-x-10))+C
=sqrt((5-x)(5+x)) + 10 arctan(sqrt((5-x)(5+x))/(-x-5))+C
= sqrt(25-x^2) + 10 arctan(sqrt(25-x^2)/(-x-5))+C