Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 35

For the given integral: int 2x/(x^2+6x+13) dx , we may apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int 2x/(x^2+6x+13) dx =2 int x/(x^2+6x+13) dx
To be able to evaluate this, we apply completing the square on x^2+6x+13 .
The x^2+6x+13 resembles ax^2+bx+c where:
a= 1 and b =6 that we can plug-into (-b/(2a))^2 .
(-b/(2a))^2= (-(6)/(2*1))^2
= (-6/2)^2
= (-3)^2
=9
To complete the square, we add and subtract 9:
x^2+6x+13 +9 -9
Group them as: (x^2+6x+9)-9+13
Simplify: (x^2+6x+9)+4
Apply factoring for the perfect square trinomial: x^2+6x+9 = (x+3)^2
(x^2+6x+9)+4=(x+3)^2 + 4
Which means x^2+6x+13 =(x+3)^2 + 4 then the integral becomes:
2 int x/sqrt(x^2+6x+13) dx =2 int x/((x+3)^2 + 4) dx
For the integral part, we apply u-substitution by letting:
u = x+3 then x= u-3 and du =dx
Then,
2 int x/((x+3)^2 + 4) dx= 2 int (u-3)/(u^2 + 4) du
Apply the basic integration property: : int (u+v) dx = int (u) dx + int (v) dx .
2 int (u-3)/(u^2 + 4) du=2 [int u/(u^2 + 4) du - int 3/(u^2 + 4) du]

For the integration of int u/(u^2 + 4) du , let:
v=u^2+4 then dv =2u du or (dv)/2 = u du .
Then,
int u/(u^2 + 4) du = int ((dv)/2)/(v)
= 1/2 int (dv)/(v)
= 1/2ln|v|+C
Plug-in v= u^2+4, we get: int u/(u^2 + 4) du =1/2ln|u^2+4|+C
For the second integration: - int 3/(u^2 + 4) du , we follow the basic integration formula for inverse tangent function:
int (du)/(u^2+a^2) = 1/a arctan(u/a)+C
Then,
- int 3/(u^2 + 4) du =-3 int (du)/(u^2 + 2^2)
= -3 *1/2arctan(u/2)+C
=-3/2 arctan(u/2)+C
Combine the results, we get:
2 [int (u/(u^2 + 4) du - int 3/(u^2 + 4) du]
=2*[ 1/2ln|u^2+4|-3/2arctan(u/2)]+C
= ln|u^2+4| - 3arctan(u/2)+C
Plug-in u=x+3 to solve for the final answer:
int 2x/(x^2+6x+13) dx= ln|(x+3)^2+4| - 3arctan((x+3)/2)+C