Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 34

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
Thenx^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)