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