Find ∫(x+3)√(34x-x²)dx

∫(x+3)√(34x-x²)dx=∫(x+2)√(34x-x²)dx+∫√(34x-x²)dx=∫(x+2)√7-(x+2)²dx+∫√7-(x+2)²dx
Substituting x+2=√7sinθ ⇒dx=√7cosθdθ, we get:
∫(x+2)√7-(x+2)²dx+∫√7-(x+2)²dx=∫√7sinθ√7cos²θ(√7cosθdθ)+∫√7cosθ(√7cosθdθ)=7∫sinθcos²θdθ+7∫cos²θdθ=
7(1/3)cos³θ+7/2sin2θ+C
Substituting back, we get:
7/3(1-(x+2/√7)²)³/²+7/2(x+2)²+C=-(7-(x+2)²)³/²(3)+7/2(x+2)²+C