#5-x²-4x = 9 - (x + 2)^2#
So the integral is:
#int \ (x+3)/(sqrt (9 - (x + 2)^2)) \ dx#
#int \ (x+2)/(sqrt (9 - (x + 2)^2)) + \ (1)/(sqrt (9 - (x + 2)^2)) \ dx#
This is handy on 2 counts.
For the first part: note that: #d/(dx) ( sqrt (9 - (x + 2)^2)) = 1/2 * 1/( sqrt (9 - (x + 2)^2)) * -2(x+2) = - (x+2)/( sqrt (9 - (x + 2)^2)) #
And so:
#int \ (x+2)/(sqrt (9 - (x + 2)^2)) = - int \ d ( sqrt (9 - (x + 2)^2)) = - sqrt (9 - (x + 2)^2) + C = - sqrt( 5 - x^2 - 4x ) +C#
For the second part, if we let #x + 2 = 3 sin \ omega#, then we have:
#int \ \ (1)/(sqrt (9 - 9 sin^2 \ omega))\ \ d( 3 sin \ omega - 2)#
#int \ \ (3cos omega )/(3cos omega)\ \ d omega#
#= omega + C#
# = sin^(-1) ((x + 2)/3) + C#
Combining these:
#I = - sqrt( 5 - x^2 - 4x ) + sin^(-1) ((x + 2)/3) + C#
