Answer
$2x^2-4\sqrt{16-x^2}+C$
Work Step by Step
$\frac{dy}{dx}=4x+\frac{4x}{\sqrt{16-x^2}}$
$\int\frac{dy}{dx}dx=\int\left(4x+\frac{4x}{\sqrt{16-x^2}}\right)dx=\int 4xdx+\int\frac{4x}{\sqrt{16-x^2}}dx$
$=2x^2+\int\frac{4x}{\sqrt{16-x^2}}dx$
Let $u=16-x^2\hspace{8mm}du=-2xdx\hspace{8mm}\frac{du}{-2x}=dx$
Therefore
$\int\frac{dy}{dx}dx=2x^2+\int\frac{4x}{\sqrt{16-x^2}}dx=2x^2+\int\frac{4x}{\sqrt{u}}\frac{du}{-2x}=2x^2-2\int\frac{1}{\sqrt{u}}du$
$2x^2-2\int u^{-\frac{1}{2}}du=2x^2-2(2u^{\frac{1}{2}})+C=2x^2-4\sqrt{16-x^2}+C$
