Find :

(x + 3) √(3 - 4x - x²) dx =
add and subtract 4 from above equation:
= ∫ (x + 3) √(3 + 4 - 4 - 4x - x²) dx
= ∫ (x + 3) √[7 - (x² + 4x + 4)] dx
= ∫ (x + 3) √[7 - (x + 2)²] dx
= ∫ (x + 3) √[(√7)² - (x + 2)²] dx
Let:
x + 2 = (√7)sinθ
x = (√7)sinθ - 2
dx = (√7)cosθ. dθ
yielding, by trig substitution:
∫ (x + 3) √[(√7)² - (x + 2)²] dx

= ∫ {[(√7)sinθ - 2] + 3} √{(√7)² - [(√7)sinθ]²} dx

= ∫ [(√7)sinθ - 2 + 3] [√(7 - 7sin²θ)] (√7)cosθ dθ

= ∫ [(√7)sinθ + 1] {√[7(1 - sin²θ)]} (√7)cosθ dθ

(replacing 1 - sin²θ with cos²θ)
= ∫ [(√7)sinθ + 1] [√(7cos²θ)] (√7)cosθ dθ

= ∫ [(√7)sinθ + 1] (√7)cosθ (√7)cosθ dθ

= ∫ [(√7)sinθ + 1] (√7)²cos²θ dθ

= ∫ [(√7)sinθ + 1] 7cos²θ dθ

= ∫ [7(√7)cos²θ sinθ + 7cos²θ] dθ

Rewrite the argument of the second cos²θ as (2θ)/2:
∫ {7(√7)cos²θ sinθ + 7cos²[(2θ)/2]} dθ

= let's apply (to cos²[(2θ)/2]) the half - angle formula cos²(α/2) = (1 + cos α)/2:
∫ {7(√7)cos²θ sinθ + 7{[1 + cos(2θ)] /2} } dθ

= ∫ {7(√7)cos²θ sinθ + (7/2)[1 + cos(2θ)]} dθ

= ∫ [7(√7)cos²θ sinθ + (7/2) + (7/2)cos(2θ)] dθ

= 7(√7) ∫ cos²θ sinθ dθ + (7/2) ∫ dθ + (7/2) ∫ cos(2θ) dθ
(dividing and multiplying the last integral by 2 that is the derivative of the argument 2θ)
= 7(√7) ∫ cos²θ sinθ dθ + (7/2) ∫ dθ + (7/2)(1/2) ∫ cos(2θ) 2 dθ = 7(√7) ∫ cos²θ sinθ dθ + (7/2)θ + (7/4) ∫ cos(2θ) d(2θ)
(∫ cos[f(x)] d[f(x)] = sin[f(x)] + C)
= 7(√7) ∫ cos²θ sinθ dθ + (7/2)θ + (7/4)sin(2θ)
let's change signs in the remaining integrand:
= 7(√7) ∫ ( - cos²θ) ( - sinθ) dθ + (7/2)θ + (7/4)sin(2θ) =
(being - sinθ the derivative of cosθ)
= 7(√7) ∫ ( - cos²θ) d(cosθ) + (7/2)θ + (7/4)sin(2θ)
= - 7(√7) ∫ cos²θ d(cosθ) + (7/2)θ + (7/4)sin(2θ)
(by the integration rule ∫ f(x)ⁿ d[f(x)] = [1/(n + 1)] f(x)ⁿ⁺¹ + C)
= - 7(√7) [1/(2 + 1)] (cosθ)²⁺ ¹ + (7/2)θ + (7/4)sin(2θ) + C
= - 7(√7) (1/3)(cosθ)³ + (7/2)θ + (7/4)sin(2θ) + C
(applying the double - angle formula sin(2θ) = 2sinθ cosθ)
= - [(7√7)/3]cos³θ + (7/2)θ + (7/4)2sinθ cosθ + C =
= - [(7√7)/3]cos³θ + (7/2)θ + (7/2)sinθ cosθ + C
But x + 2 = (√7)sinθ; hence:
sinθ = (x + 2)/√7
θ = arcsin[(x + 2)/√7]
cosθ = √(1 - sin²θ)

= √{1 - [(x + 2)/√7]²}

= √{1 - [(x + 2)²/7]}

= √{1 - [(x² + 4x + 4)/7]}

= √{[7 - (x² + 4x + 4)} /7}
= √[(7 - x² - 4x - 4)} /7]
= [√(3 - 4x - x²)] /√7
then, substituting back:
- [(7√7)/3]cos³θ + (7/2)θ + (7/2)sinθ cosθ + C

= - [(7√7)/3] {[√(3 - 4x - x²)] /√7}³ +
(7/2)arcsin[(x + 2)/√7] + (7/2) [(x + 2)/√7] {[√(3 - 4x - x²)] /√7} + C
= - [(7√7)/3] {[√(3 - 4x - x²)³] /√7³} + (7/2)arcsin[(x + 2)/√7] + (7/2) [(x + 2)/√7²] √(3 - 4x - x²) + C

= - [(7√7)/3] {[√(3 - 4x - x²)³] /√(7² ∙ 7)} + (7/2)arcsin[(x + 2)/√7] + (7/2)[(x + 2)/7] √(3 - 4x - x²) + C

= - [(7√7)/3] {[√(3 - 4x - x²)³] /(7√7)} + (7/2)arcsin[(x + 2)/√7] + (1/2)(x + 2)√(3 - 4x - x²) + C

ending with:
= - (1/3)√(3 - 4x - x²)³ + (7/2)arcsin[(x + 2)/√7] + (1/2)(x + 2)√(3 - 4x - x²) + C.