`int_-2^2 dx/(x^2+4x+13)` Find or evaluate the integral by completing the square

 To evaluate the given integral: `int_(-2)^(2)(dx)/(x^2+4x+13)` ,

 we follow the first fundamental theorem of calculus: 

If f is continuous on closed interval [a,b], we follow:

` int_a^bf(x)dx = F(b) - F(a)`

 where F is the anti-derivative or indefinite integral of f on closed interval `[a,b]` .

  To determine the `F(x)` , we apply completing the square on the trinomial: `x^2+4x+13.`

Completing the square:

`x^2+4x+13 ` is in a form of   ` ax^2 +bx+c`

 where:

...

 To evaluate the given integral: `int_(-2)^(2)(dx)/(x^2+4x+13)` ,

 we follow the first fundamental theorem of calculus: 

If f is continuous on closed interval [a,b], we follow:

` int_a^bf(x)dx = F(b) - F(a)`

 where F is the anti-derivative or indefinite integral of f on closed interval `[a,b]` .

  To determine the `F(x)` , we apply completing the square on the trinomial: `x^2+4x+13.`

Completing the square:

`x^2+4x+13 ` is in a form of   ` ax^2 +bx+c`

 where:

a =1

b =4

 c= 13

 To complete square ,we add and subtract `(-b/(2a))^2` on both sides:

With a=1 and b = 4 then:

`(-b/(2a))^2 =(-4/(2*1))^2 = 4`

Then`x^2+4x+13` becomes:

`x^2+4x+ 13 +4-4`

`(x^2+4x+4) + 13 -4`

Applying` x^2 +4x +13 =(x+2)^2 + 9 ` in the given integral, we get:

`int_(-2)^(2) (dx)/(x^2+4x+13) =int_(-2)^(2) (dx)/((x+2)^2 + 9)`

 The integral form: `int_(-2)^(2) (dx)/((x+2)^2 + 9) ` resembles the 

basic integration formula for inverse tangent function:

`int_a^b (du)/(u^2+c^2) = (1/c)arctan(u/c) |_a^b`

Using u-substitution, we let `u = x+2 ` then `du = 1dx`  or  ` du=dx.`

where the boundary  limits:  upper bound = 2 and lower bound =-2

and  `c^2 = 9` then `c = 3`

The indefinite integral will be:

`int_(-2)^(2) (dx)/((x+2)^2 + 9) =int_(-2)^(2) (du)/(u^2 + 9)`

            `=(1/3)arctan(u/3) |_(-2)^(2)`

Plug-in` u=x+2` to solve for` F(x)` :

`(1/3)arctan(u/3) |_(-2)^(2)=(1/3)arctan((x+2)/3) |_(-2)^(2).`

We now have  `F(x)|_a^b=(1/3)arctan(u/3) |_(-2)^(2).`

Applying `F(x)|_a^b= F(b)-F(a)` , we get:

`(1/3)arctan(x+2/3) |_(-2)^(2)`

 ` =(1/3)arctan((2+2)/3) -(1/3)arctan((-2+2)/3)`

` =(1/3)arctan(4/3) -(1/3)arctan(0/3)`

`=(1/3)arctan(4/3) -0`

  `=(1/3)arctan(4/3)`