# `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))dx`resembles one of the formulas from the integration table. It follows the integration formula for rational...

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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))dx`resembles 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 `

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