Let x=2\sin\theta\implies dx=2\cos\theta\ d\theta
\therefore \int (x+3)\sqrt{4-x^2}\ dx
=\int (2\sin\theta+3)\sqrt{4-4sin^2\theta}\ (2\cos\theta\ d\theta)
=\int (2\sin\theta+3)(2\cos\theta)\ (2\cos\theta\ d\theta)
=4\int (2\sin\theta+3)\cos^2\theta\ d\theta
=4\int (2\cos^2\theta \sin\theta+3\cos^2\theta)\ d\theta
=8\int \cos^2\theta \sin\theta\ \d\ theta+12\int \cos^2\theta\ d\theta
=-8\int \cos^2\theta(- \sin\theta\ \d\ theta)+12\int \frac{ 1+\cos2\theta}{2}\ d\theta
=-8\int \cos^2\theta \ d(\cos\theta)+6\int (1+\cos2\theta)\ d\theta
=-8(\frac{\cos^3\theta}{3})+6(\theta+{\sin2\theta}/2)+C
=-8/3\cos^3\theta+6\theta+3\sin2\theta+C
=6\sin\theta\cos\theta-8/3\cos^3\theta+6\theta+C
=6(x/2)\sqrt{1-(x/2)^2}-8/3(1-(x/2)^2)^{3/2}+6\sin^{-1}(x/2)+C
=3/2x\sqrt{4-x^2}-1/3(4-x^2)^{3/2}+6\sin^{-1}(x/2)+C