Answer
$2\pi$
Work Step by Step
$\int\frac{8\,dx}{x^{2}-2x+2}=8\int\frac{\,dx}{x^{2}-2x+2}=8\int\frac{\,dx}{(x-1)^{2}+1}$
Substituting $u=x-1$ so that $dx=du$, we have
$8\int\frac{\,dx}{(x-1)^{2}+1}=8\int\frac{du}{u^{2}+1}=8\tan^{-1}(u)+C$
$=8\tan^{-1}(x-1)+C$
Therefore $\int^{2}_{1}\frac{8\,dx}{x^{2}-2x+2}=[8\tan^{-1}(x-1)]^{2}_{1}$
$=8\times\tan^{-1}(1)-8\times\tan^{-1}(0)=8\times\frac{\pi}{4}-8\times0=2\pi$
