Start from the given
int (2+x)/sqrt(4-2x-x^2)dx
Start with Algebra by completing the square
4-2x-x^2=-(x^2+2x-4)=-(x^2+2x+1-1-4)
and
4-2x-x^2=-((x+1)^2-5)=5-(x+1)^2
then
int (2+x)/sqrt(4-2x-x^2)dx=int (2+x)/sqrt(5-(x+1)^2)dx
The Trigonometric Substitution
Let x+1=sqrt(5)*sin theta
and x=sqrt(5)*sin theta -1
and dx=sqrt(5)*cos d theta
Let's do the substitution
int (2+x)/sqrt(5-(x+1)^2)dx=
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/sqrt(5-(sqrt(5)*sin theta)^2)
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/sqrt(5-5*sin^2 theta)
continue simplification by trigonometric identities
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*sqrt(1-sin^2 theta))
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*sqrt(cos^2 theta))
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*cos theta)
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*cos theta)
and
int (2+x)/sqrt(5-(x+1)^2)dx=int(sqrt(5)sin theta + 1)d theta
int (2+x)/sqrt(5-(x+1)^2)dx=int(1+sqrt(5)sin theta )d theta
int (2+x)/sqrt(5-(x+1)^2)dx=theta-sqrt(5)*cos theta+C
Now, time to imagine your right triangle with
angle theta
Let x+1 the Opposite side to angle theta
Let sqrt5 the Hypotenuse
Let sqrt(4-2x-x^2) the Adjacent side to angle theta
Return the variables
int (2+x)/sqrt(5-(x+1)^2)dx=theta-sqrt(5)*cos theta+C
int (2+x)/sqrt(5-(x+1)^2)dx=arcsin ((x+1)/(sqrt5))-sqrt(4-2x-x^2)+C
I hope the explanation is useful....God bless...