∫√(5-4x-x^2)dx
=∫√[9-(x+2)^2]dx
x+2=3sinu dx=d(x+2)=d3sinu=3cosudu
=∫9cosu^2du
=(9/2)∫(1+cos2u)du
=(9/2)u+(9/4)sin2u+C
=(9/2)u+(9/2)sinucosu+C
=(9/2)arcsin[(x+2)/3] +(9/2)[(x+2)/3]√(1-(x+2)^2/9) +C
=(9/2)arcsin[(x+2)/3] +(1/2)(x+2)√(5-4x-x^2) +C