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...
