Find :∫(x+3)√(3-4x-x²)dx

I=∫(x+3)√(3-4x-x²)dx
Put x+3=p
d/dx(3-4x-x²)+q ⇒x+3=p (-4-2x)+q
Therefore,-2p=1 ⇒p=-½
-4p+q=3 ⇒q=1
So given integral,
I=∫(-½(-4-2x)+1)√(3-4x-x²)dx
⇒I=∫-½ (-4-2x)√(3-4x-x²)dx+∫√(3-4x-x²)dx
Put 3-4x-x²=t ⇒(-4-2x)dx=dt
I=-½∫√tdt+∫√(7-(x+2)²)dx
I=-½.⅔.t³/²+((x+2)/2)√(7-(x+2)²)+7/2.sin⁻¹((x+2)/√7)+C
⇒I=-⅓ (3-4x-x²)³/²+((x+2)/2)√(3-4x-x²)+7/2sin⁻¹((x+2)/√7)+C