To begin with, your "du" is incorrect. Taking $u= \sqrt{1- x^2}= (1- x^2)^{1/2}$, $du= (1/2)(1- x^2)^{-1/2}(-2x)= -\frac{x}{\sqrt{1- x^2}}$. You did not use the "chain rule".
Addition by a "clever zero" means adding something like $1-1$. It adds to zero. So, for instance, inside the square root, you can add $-4x+4x$. Then you have:
$\sqrt{4-x^2} = \sqrt{4-4x-x^2+4x} = \sqrt{(2-x)^2+4x}$
This is not the "clever zero" the problem was referring to and in fact makes the problem harder, but it is an example of adding a zero to change the problem. You just need to figure out the zero needed to make the problem easier rather than harder.
To check you work, let's solve the integral another way. Then, you can check to see if you get the same solution using integration by parts. How about the substitution $x=2\sin \theta$? Then you have $dx = 2\cos \theta d\theta$. Your integral becomes:
$\displaystyle 2\int \sqrt{4(1-\sin^2 \theta)}\cos \theta d\theta = 2\int 2\cos^2\theta d\theta = 2\int \left(\cos(2\theta)+1\right)d\theta = \sin(2\theta)+2\theta + C = 2\arcsin\left(\dfrac{x}{2}\right)+\dfrac{x\sqrt{4-x^2}}{2}+C$