int x/sqrt(x^2+6x+12) dx Complete the square and find the indefinite integral

Recall indefinite integral follows int f(x) dx = F(x)+C
 where:
f(x) as the integrand
F(x) as the antiderivative of f(x)
C as the constant of integration.
To evaluate the given integral:  int x/sqrt(x^2+6x+12)dx , we  may apply completing the square at the trinomial: x^2+6x+12 .
Completing the square:
x^2+6x+12 is in a form of ax^2 +bx+c
 where:
a =1
b =6
 c= 12
 To complete square ,we add and subtract (-b/(2a))^2 :
With a=1 and b = 6 then:
(-b/(2a))^2 =(-6/(2*1))^2 = 9
Then x^2+6x+12 becomes:
x^2+6x+ 12 +9-9
(x^2+6x+9) + 12 -9
(x+3)^2 +3
Applying x^2 +6x +12 =(x+3)^2 + 3 in the given integral, we get:
int x/sqrt(x^2+6x+12)dx=int x/sqrt((x+3)^2 + 3)dx
We may apply u-substitution by letting:  u = x+3 or x =u-3 then du = dx .
The integral becomes:
int x/sqrt((x+3)^2 + 3)dx =int (u-3)/sqrt(u^2 + 3)du
Apply the basic integration property: int (u-v) dx = int (u) dx - int (v) dx .
int (u-3)/sqrt(u^2 + 3)du =int u/sqrt(u^2 + 3)du -int 3/sqrt(u^2 + 3)du
For the integral of  int u/sqrt(u^2 + 3)du , we may apply formula from integration table: int u/sqrt(u^2+-a^2) du = sqrt(u^2+-a^2) +C
Take note we have + sign inside the root then we follow: int u/sqrt(u^2+a^2) du = sqrt(u^2+a^2) +C .
int u/sqrt(u^2 + 3)du=sqrt(u^2+3)                         
For the integral of int 3/sqrt(u^2 + 3)du , we use the basic integration property: int cf(x)dx = c int f(x) dx.
int 3/sqrt(u^2 + 3)du = 3int 1/sqrt(u^2 + 3)du
From integration table, we may apply the formula for rational function with roots:
int 1/sqrt(x^2+-a^2)dx = ln|x+sqrt(x^2+-a^2)| +C
With just (+) inside the root, we follow:int 1/sqrt(x^2+a^2)dx = ln|x+sqrt(x^2+a^2)| +C.
3int 1/sqrt(u^2 + 3)du=ln|u+sqrt(u^2+3)|
Combining the results, we get:
int (u-3)/sqrt(u^2 + 3)du =sqrt(u^2+3) -ln|u+sqrt(u^2+3)| +C
Plug-in u = x+3 on sqrt(u^2+3) -ln|u+sqrt(u^2+3)| +C , we get the indefinite integral as:
int x/sqrt(x^2+6x+12)dx =sqrt((x+3)^2+3) -ln|x+3+sqrt((x+3)^2+3)| +C
Recall: (x+3)^2+3 =x^2+6x+12 then the indefinite integral can also be expressed as:
int x/sqrt(x^2+6x+12)dx =sqrt(x^2+6x+12) -ln|x+3+sqrt(x^2+6x+12)| +C